OverUnity Research

I recently started posting papers on Harti's overunity.com site and got invited to join overunityresearch.com so here I am.  For some background I live in the UK, I am 80 years old and I have been following the OU scene since 1994.  My working life was spent in the UK defence industry initially as an electronics engineer but later as an electromagnetic engineer/physicist.  I had several patents (now expired) and a number of patent applications, and I am a member of the Institute of Patentees and Inventors (which doesn't mean much but I can put M. Inst. P. I. after my name, or "mince pie" as it is colloquially known  :) ).  I intend to put all my papers on OU on the web and here is the start.

I was fortunate enough to be able to do some experiments on the generator version of Marinov's Siberian Coilu, which I choose to call the Marinov Generator.  Essentally this is a slip-ring with two brushes at diametrically opposite positions.  In my simple version a magnet is placed near each brush.  When the slip-ring is rotated a DC voltage is generated across the brushes.  It is possible that this is not classical flux-cutting induction, but is a form of longitudinal induction (along the velocity direction) from movement through the vector potential, and would occur if a closed magnetic circuit were used where the external flux is zero.  If so this offers OU potential.

Here is the first of several papers, this one describing the experiment.  More to follow

Welcome to the forum, Smudge !

Regards, ION (Vortex1)

Here are some more of my papers.  Enjoy!

I wonder if it would be possible to design an experiment where torque or current draw of the driving motor is measured with the Marinov generator open circuit and short circuit to attempt to detect loading effects, if any.

This would have to require a lightweight, sensitive apparatus with a minimum of friction.

As far as extracting energy from this low voltage high current generator, I would use a large inductor with low resistance windings to store the current in the magnetic field, then release it in a timed manner from the back emf which is now a current source, and can charge a capacitor through a diode to any desired voltage, depending on the capacitor value and the inductance. (flyback converter)

Welcome Smudge

Thanks for sharing your work here  O0

I am also in the UK, South East coast.

Regards
Peter

Here are some PDF links extracted from http://www.overunity.com/15083/the-new-generator-no-effect-counter-b-emf-part-2-selfrunning/msg424135/#msg424135 accidental post in wrong thread:

Deutsche Physik 1992 and 1996:
http://www.beinsa.info/index.php?option=com_jdownloads&Itemid=16&view=finish&cid=4366&catid=73
http://www.beinsa.info/index.php?option=com_jdownloads&Itemid=16&view=finish&cid=4352&catid=71
http://www.beinsa.info/index.php?option=com_jdownloads&Itemid=16&view=finish&cid=4337&catid=71

Also there is thread in overunity.com forum:

http://www.overunity.com/14691/the-marinov-generator/

Thanks Smudge and welcome. I'm looking forward to reading your work. Thanks for making it available.

Going away for a few days so I thought I'd get this thread going again.  Of all the OU devices I've been involved with this is the only one where there is a plausible explanation for where the energy comes from.  Also I found that I am not the only one to have performed the slip-ring Marinov generator experiment, Stan Zuala got 2mV at 2000rpm, picture of his set up below.

What i would like to see is an experiment that shows DC induction into a coil, see image.  It needs a slip-ring driven at high speed with diametrically opposite brushes.  Current is passed across the slip ring from a DC source.  A multi turn coil is placed close to one of the brushes where, according to my math, it should get a DC voltage induced.  There will also be slip-ring noise that will need filtering out.  The coil needs as many turns as possible using fine gauge wire.  If someone could validate this it would pave the way for more extensive research into the phenomenon because if you can create DC  E field vortices they can be used to "load" the electron spins and orbits responsible for magnetization, and it is a route to extracting energy from a PM.

Anyone up for a try?  You would get some kudos for being the first person to demonstrate this hole in our knowledge, it would be worthy of a publication in a scientific journal.

Smudge

Dear Smudge.

If I have correctly interpreted your drawing it is an insulated disc that carries a Copper ring driven by an electric motor ?

Should be fairly simple to construct can the Copper ring have a joint ? Could I Silver solder said joint ?

Cheers Grum.

Quote from: Smudge on 2015.11.06, 20:05:35
Going away for a few days so I thought I'd get this thread going again.  Of all the OU devices I've been involved with this is the only one where there is a plausible explanation for where the energy comes from.  Also I found that I am not the only one to have performed the slip-ring Marinov generator experiment, Stan Zuala got 2mV at 2000rpm, picture of his set up below.

What i would like to see is an experiment that shows DC induction into a coil, see image.  It needs a slip-ring driven at high speed with diametrically opposite brushes.  Current is passed across the slip ring from a DC source.  A multi turn coil is placed close to one of the brushes where, according to my math, it should get a DC voltage induced.  There will also be slip-ring noise that will need filtering out.  The coil needs as many turns as possible using fine gauge wire.  If someone could validate this it would pave the way for more extensive research into the phenomenon because if you can create DC  E field vortices they can be used to "load" the electron spins and orbits responsible for magnetization, and it is a route to extracting energy from a PM.

Anyone up for a try?  You would get some kudos for being the first person to demonstrate this hole in our knowledge, it would be worthy of a publication in a scientific journal.

Smudge

In one of Tesla's methods for conversion, we find him shorting a coil where AC is being induced on itself, doing so results in the production of unidirectional impulses, high voltage DC.  Fast forward and we see parallels in coil shorting experiments.  Through this method, DC is produced simply and effectively, and most importantly, without the need for proper rectification.  The problem with this method of conversion of AC to DC (off topic I know, however, I feel its related) is in all the familiar geometries, the secondary induced magnetic field interacts antagonistically with the inducing field.  When the circuit controller is activated, a  massive current is induced and Lenz works his magic.  The solution to this problem takes the form of what I refer to as a asymmetrical, mono-polar (the term mono-polar here refers to the utilization of both halves of the dipole) magnetic field.  This modification in geometry is applied to both inducing and induced circuits.  The result is dramatically reduced drag in all generation applications, but more importantly, in those circuits where unidirectional impulse currents are desired.

Please forgive my intrusion.


Regards

I posted a paper on using the Earth's magnetic vector potential over on PhysicsProf's thread.  If I am right then the experiment shown in the image here should work.  It simply involves a slip ring being rotated in the Earth's vector potential and having brushes diametrically opposite each other.  The diagonal joining the brushes must be at right angles to the vector potential that runs E-W.  If the theory stacks up there should be a DC voltage across the brushes.  Swinging the thing through 180 degrees or changing the rotation direction should cause a reversal of the voltage.  Seems a simple method to prove or disprove the theory.

Smudge

Quote from: Smudge on 2016.01.28, 19:55:31
I posted a paper on using the Earth's magnetic vector potential over on PhysicsProf's thread.  If I am right then the experiment shown in the image here should work.  It simply involves a slip ring being rotated in the Earth's vector potential and having brushes diametrically opposite each other.  The diagonal joining the brushes must be at right angles to the vector potential that runs E-W.  If the theory stacks up there should be a DC voltage across the brushes.  Swinging the thing through 180 degrees or changing the rotation direction should cause a reversal of the voltage.  Seems a simple method to prove or disprove the theory.

Smudge

Dear Smudge.

It's been a while but did you miss my post a couple up from this one?

Cheers Grum.

Quote from: Grumage on 2016.01.28, 20:04:15
Dear Smudge.

It's been a while but did you miss my post a couple up from this one?

Cheers Grum.

Sorry Grum, I did miss your post.  Yes the slip-ring is just a copper hoop on an insulated disc.  And it could be fabricated from copper tape with the ends silver soldered.  If you do make one then you could try this earth field experiment as well as replicating the Marinov generator experiment that I did where you place magnets close to the brushes.

While on this earth field stuff it strikes me that there could be an explanation for the weird machine in that Swiss commune that looks a bit like a Wimshurst machine.  The manner in which the grad(v.A) force field works on conduction electrons in a spinning disc is a little uncertain, but it could result in anomalous surface charge that could be bled off into capacitors (Leydon jars in that weird machine).

Smudge

Quote from: Smudge on 2016.01.30, 10:20:06


Sorry Grum, I did miss your post.  Yes the slip-ring is just a copper hoop on an insulated disc.  And it could be fabricated from copper tape with the ends silver soldered.  If you do make one then you could try this earth field experiment as well as replicating the Marinov generator experiment that I did where you place magnets close to the brushes.

While on this earth field stuff it strikes me that there could be an explanation for the weird machine in that Swiss commune that looks a bit like a Wimshurst machine.  The manner in which the grad(v.A) force field works on conduction electrons in a spinning disc is a little uncertain, but it could result in anomalous surface charge that could be bled off into capacitors (Leydon jars in that weird machine).

Smudge

Dear Smudge.

I have an amazing stock of resources in my workshop, could you please provide a drawing of the apparatus required?

I can make a disc up to 14" in dia out of Polycarbonate sheet, various thicknesses up to 3/8" I also have Polypropylene and PVC sheet.

Copper sheets in 0.2 and 0.3 thickness.

I await a plan!     ;)

Cheers Grum.

Quote from: Grumage on 2016.01.30, 15:06:28
Dear Smudge.

I have an amazing stock of resources in my workshop, could you please provide a drawing of the apparatus required?

I can make a disc up to 14" in dia out of Polycarbonate sheet, various thicknesses up to 3/8" I also have Polypropylene and PVC sheet.

Copper sheets in 0.2 and 0.3 thickness.

I await a plan!     ;)

Cheers Grum.

Having got the little grey cells going on the Testatika thingy if you can wait a little while I will come up with a proof of concept experiment.
Smudge

I have decided to reactivate this bench.  Over the last two years I have looked at many texts dealing with the derivation of our electromagnetic laws and in particular those that use as a starting point the so-called hidden momentum or electro-kinetic momentum of a point charge within a magnetic vector potential.  In a nutshell this states that an electron of charge e within a vector potential A (bold character represents a vector) has a hidden (i.e. non-mechanical, not related to velocity) momentum eA.   To many schooled in classical mechanics the idea of a body possessing momentum that is not related to its velocity is hard to grasp.  However it should be realized that electro-dynamic or electro-kinetic forces are transmitted to a body via photons or sub-photons, and these invisible particles travelling at light velocity do carry momentum.  Thus the so-called hidden momentum could arise from some form of supplied momentum, supplied by interaction with those invisible space particles.

The upshot of this momentum approach is the derivation of our three well-known classical forces, (a) the Coulomb force expressed by the electric field E=-grad(phi) where phi is the potential from nearby charges, (b) transformer induction expressed as volts per turn V=-d(Phi)/dt  where Phi is the magnetic flux enclosed by the turn and (c) motional or flux-cutting induction expressed as E=vXB , this latter being taught by the use of Fleming's LH and RH rules.  But the approach also brings up a third form of induction (d), an electric field that is the gradient of a scalar potential formed from the product of the tangential component of A along the velocity direction, expressed as E_A=-grad_A(v.A)  but with some components removed.  This restriction is usually expressed by stating where the gradient operates on A only and not on v (hence the subscript _A in that formula).  What those authors then missed is a rather important aspect of this new induction.  All text books will tell you that the closed line integral of something that is the gradient of a scalar is zero, and it is then assumed that this applies to the new term, and that says it is impossible to induce a voltage into a closed circuit and the Marinov generator will not work.  The books then go on to say that the closed line integral of any E field can have only two possible voltage values, either (a) equal to the rate of change of magnetic flux passing through that closed line or (b) zero if there is no enclosed magnetic flux or if the flux is constant. 

I have looked carefully at the derivations by examining every component of the vector identities to see which components of E_A=-grad_A(v.A)  have to be discarded.  This reveals that voltage can be induced into a closed circuit.  It is possible to have electrons travelling at a trivial velocity (e.g. drift velocity within a conductor) over part of the loop and at high velocity (e.g. carried in a slip-ring) over the remaining part of the loop, and then it is possible for voltage to be induced into that closed circuit.  Of course it requires a non-uniform A field to be present, such as that presented by the magnets in the Marinov generator.  Using this new induction term the Marinov generator will work.

As this is something that defies our classically taught electrodynamics I think it important to prove or disprove the existence of this new term.  It should be possible to create an experiment that produces up to 100mV DC, not the 3mV of my earlier work with BGB.  The downside of the Marinov generator is its low voltage, if it really works it is a low voltage high current machine.  The upside is that it should be overunity, its power is not taken from the drive shafts but comes from the quantum dynamos (the electron circulations) within the magnets.  It seems possible to combine an OU Marinov generator mechanically driven by, and electrically connected to, a non-OU homopolar motor, perhaps using rolling contacts to avoid brush friction, resulting in a free running machine.
(Edit minor changes)

Smudge

Forget my earlier posts about using the earth's magnetic field, the induction I was expecting won't occur.  Here is an image of the Marinov motor taken from another document but modified to place a load resistor where the drive battery was.  This basically depicts the Marinov generator (but ignore the current and/or the rotation directions, they will be wrong for the generator).  It requires a slip ring within which the magnets can be placed, so it can't be a copper hoop on the outer edge of a plastic disc.  If anyone has an old toroidal winding machine that can be modified, that could be used since it drives a circular shuttle that could become the slip ring.
Smudge

Your paper from 2009.

Hello Smudge
What is the difference between this new proposal and the BGB Engineering Ltd experiment you performed?
Rgds
Cortazar

Hi Cortazar,

At BGB I only had access to one of their standard slip rings so it was not possible to have anything inside it.  I just had some disc magnets placed close to the brushes.  The slip ring was therefore within a magnetic B field and that raises doubt about the modus operandi.  For a real test we require a closed magnetic circuit like a magnetized ring core or two parallel bar magnets with keepers across the ends so that there is only the magnetic vector potential A field around it, i.e. no B field outside the magnet.  And that magnet has to be within the slip ring so it requires a clever mechanical arrangement to drive and support the ring.

Smudge

If I had the time I would build this maybe later, so I started with a preliminary drawing. All minor details of the drawing will be added later as time permits but should be easy for the moderately skilled mechanical builder.

Version 2 allows torque indication on the magnet assembly as it is free to pivot against a spring.

Regards

Here's how I did it, for the motor version. A triaxial axis was arranged so that the stator ring, the magnet assembly and the brush/power supply assembly could all rotate independently about the vertical axis of the assembly.

I used another different test rig but with the same co-axial suspension idea to explore the generator version.

A large copper vacuum flange gasket is ideal for the "slip ring" or stator.

Brushes should be rigged so that they can contact the stator ring either from outside edge or inside edge. This is very important!

Also, allow fringing fields from the magnet rotor assembly. If you use "keepers" make sure that you can also test without them.

There is a lot to be learned from this little gadget and its variations. Is it "merely" a homopolar motor/generator with unusual geometry, or is there really something to Marinov's claims about the vector potential "field"?

Thanks ION and TK for your interest in this.  I have researched a number of technical papers that all get to that truncated gradient of the electro-kinetic potential as an additional term to be added to the classical ones.  I'll try to precis all that information and publish it here.  What all those authors have missed is the fact that the truncation, the elimination of all terms involving velocity changes, gives that term a special characteristic in that a closed line integral (the emf induced into a closed circuit) is not necessarily zero or null.  Only in the case where the speed of the electrons are constant is the emf zero, so for a normal closed electrical circuit where drift speeds are negligible you will get close to a null.  But with the slip-ring generator the electrons (and ions) in the slip ring are not travelling at trivial velocity, and our closed electrical circuit carrying load current has two regions (the brush tips) where there is a sudden jump in electron velocity.  Electrons that are suddenly accelerated radiate an electric field and I have shown that field to have unusual characteristics; taken one electron at a time I have shown if the jump occurs where there is an A field then the radiated E field will apply a load to the electric current circulations that supply that A field.  There is therefore a good reason for the Marinov generator to be OU with the energy being taken from the atomic magnetic dipoles within the magnets.  Whether this has a cooling effect on the magnets is yet to be seen, but I think there is a link to the virtual particles that comprise the aether and that could be the energy source. 

With various people having experimented with the Marinov motor, I am surprised that no one seems to have played with the generator version.  The only evidence I can find is the Distinti Paradox 2 experiment.  One pointer to this being possible is the Solunin-Kostin experiment where he placed a magnetized ring core around the neck of a CRT near the cathode, and demonstrated that the A field could create a deflection of the beam.

Smudge

Quote from: ION on 2018.05.01, 21:20:40
If I had the time I would build this maybe later, so I started with a preliminary drawing. All minor details of the drawing will be added later as time permits but should be easy for the moderately skilled mechanical builder.

Version 2 allows torque indication on the magnet assembly as it is free to pivot against a spring.

Regards

The magnet assembly need not be the elongated pair of bar magnets with keepers, it could be a magnetized ring core.  Unfortunately you can't buy a magnetized ring core made of hard magnetic material where the field is all within the core as there is no demand for such a thing.  You can get circular arc segment magnets from KJ Magnetics and make a magnetized ring from them, FEMM simulation attached.

Alternatively a soft magnetic ring core magnetized by a DC energized toroidal coil.  Or if the induction is real then why not have that toroid AC energized, then the device will deliver AC.  That would allow for voltage to be stepped up to something useful by using a transformer.  Should still be OU as the energy would come from loading applied to the dipoles in the core, not from the current in the toroidal excitation coil.

Smudge

I have modified ION's image to show the use of a ring core.  To fit the image it appears elliptical.
Smudge

https://www.youtube.com/watch?v=K3IyiBDjOog

May be interesting... Experiment #21 from Nikolaev's book. Motor and generator experiment.
Unfortunately in Russian, but you can use auto generated and translated subtitles.

Regards,
-V.

It is my opinion that this Marinov's motor/generator is just a variation of a homopolar generator. 

There appears to be no interaction between the magnet and the ring, but I thin this is a misinterpretation.  Everyone argues that the magnetic field is contained in the closed ring magnet, and it is.  It is assumed that the electric field in contained in the ring and this is incorrect as the electric field extends across the ring due to the differences in potential which exists whether the ring is driven (generator) or powered (motor).  (see Jefimenko paper, page 20, figure 4, attached)

Orthogonal magnetic and electrical fields and "rotation" are the common requirements for homopolar induction with both rotating conductors and dielectrics.  Devices by Rowland, Wilson/Wilson, Roentgen, and Eichenwald all share these same requirements. (attached)

Quote from: Grumpy on 2018.05.02, 20:27:49
It is my opinion that this Marinov's motor/generator is just a variation of a homopolar generator.

Depends on your definition of "homopolar".  If homopolar means a DC generator then it certainly is.  But it is unlike the Faraday disc generator in that the conductor is not within a magnetic field so you can't use the flux-cutting rule that applies to the Faraday disc.   

QuoteThere appears to be no interaction between the magnet and the ring, but I thin this is a misinterpretation.  Everyone argues that the magnetic field is contained in the closed ring magnet, and it is.  It is assumed that the electric field in contained in the ring and this is incorrect as the electric field extends across the ring due to the differences in potential which exists whether the ring is driven (generator) or powered (motor).  (see Jefimenko paper, page 20, figure 4, attached)

But in those Jefimenko demonstrations the electric potential is supplied from an outside source.  In the Marinov generator it is induced and it is that induction that does not obey classical rules and the whole point of the experiment is to show that. 

QuoteOrthogonal magnetic and electrical fields and "rotation" are the common requirements for homopolar induction with both rotating conductors and dielectrics.  Devices by Rowland, Wilson/Wilson, Roentgen, and Eichenwald all share these same requirements. (attached)

Those experiments use something that is rotating in either an electric field or a magnetic field.  In the Marinov generator the slip ring is rotating in neither of those fields.  It is rotating in a constant magnetic vector potential field and that is the whole point, it is different.  Current teachings tell us that the only measurable effect of movement in such a field is the Aharanov-Bohm effect which is a quantum interference effect and not an induced voltage effect.  However the AB effect is adequately explained by the induction that hopefully will be demonstrated here.  And if it truly exists it opens a new paradigm for overunity research.
Smudge

Quote from: Smudge on 2018.05.02, 14:47:22
The magnet assembly need not be the elongated pair of bar magnets with keepers, it could be a magnetized ring core.  Unfortunately you can't buy a magnetized ring core made of hard magnetic material where the field is all within the core as there is no demand for such a thing.  You can get circular arc segment magnets from KJ Magnetics and make a magnetized ring from them, FEMM simulation attached.

Alternatively a soft magnetic ring core magnetized by a DC energized toroidal coil.  Or if the induction is real then why not have that toroid AC energized, then the device will deliver AC.  That would allow for voltage to be stepped up to something useful by using a transformer.  Should still be OU as the energy would come from loading applied to the dipoles in the core, not from the current in the toroidal excitation coil.

Smudge

Could you not just use two horse shoe magnets joined together?.
Would be a cheap way to achieve what you need.


Brad

Budget setup-if of any use.


Brad

Quote from: Smudge on 2018.05.03, 08:21:14
Depends on your definition of "homopolar".  If homopolar means a DC generator then it certainly is.  But it is unlike the Faraday disc generator in that the conductor is not within a magnetic field so you can't use the flux-cutting rule that applies to the Faraday disc.   
...
Smudge

  The Faraday generator continues to be of great interest - nearly 200 years after its invention by Michael Faraday!

   Smudge: "But it is unlike the Faraday disc generator in that the conductor is not within a magnetic field so you can't use the flux-cutting rule that applies to the Faraday disc.   "

   Let me ask this, - what if I rotate the magnet WITH the rotating Faraday disk, is there then "the flux-cutting rule"?  and - will there then be a voltage generated in this case, or not?

Quote from: Smudge on 2018.05.03, 08:21:14
Depends on your definition of "homopolar".  If homopolar means a DC generator then it certainly is.  But it is unlike the Faraday disc generator in that the conductor is not within a magnetic field so you can't use the flux-cutting rule that applies to the Faraday disc.   

There is an electric field, common to the slip ring, that cuts across the magnet.  I'm suggesting this method of induction is an electrical analog to Faraday Induction.

Quote from: Smudge on 2018.05.03, 08:21:14
But in those Jefimenko demonstrations the electric potential is supplied from an outside source.  In the Marinov generator it is induced and it is that induction that does not obey classical rules and the whole point of the experiment is to show that. 

This paper was just to show the electric field lines of a looped conductor.

Quote from: Smudge on 2018.05.03, 08:21:14
Those experiments use something that is rotating in either an electric field or a magnetic field.  In the Marinov generator the slip ring is rotating in neither of those fields.  It is rotating in a constant magnetic vector potential field and that is the whole point, it is different.  Current teachings tell us that the only measurable effect of movement in such a field is the Aharanov-Bohm effect which is a quantum interference effect and not an induced voltage effect.  However the AB effect is adequately explained by the induction that hopefully will be demonstrated here.  And if it truly exists it opens a new paradigm for overunity research.
Smudge

This was to illustrate the four similar homopolar methods that are already known, with both electric and magnetic fields.

If the induced current is caused by the  magnetic vector potential field, around the closed magnetic ring, isn't one side of this field in the opposite direction of the other side since the magnetic field vectors are in opposite directions?  I would think that this would apply to my electric field explanation as well.

There is a diagram here: https://en.wikipedia.org/wiki/Magnetic_potential

The effect is the same, either way, but how do you determine which field is producing the induction in the ring?

What if you shielded the magnet from electric fields?  Would the MVP still be there?

Sorry for the double post, but is the magnetic vector potential considered a "force"?

It's units are the same as momentum per unit charge.

Quote from: Grumpy on 2018.05.03, 14:34:27
There is an electric field, common to the slip ring, that cuts across the magnet.  I'm suggesting this method of induction is an electrical analog to Faraday Induction.

But where does your electric field come from?  If you say it is from the slip ring movement through the magnetic vector potential field then you are agreeing with this approach.

QuoteThis paper was just to show the electric field lines of a looped conductor.

Yes, a thin looped conductor connected to a voltage source.  It demonstrates the electric field from the varying voltage drop along the wire.

QuoteIf the induced current is caused by the  magnetic vector potential field, around the closed magnetic ring, isn't one side of this field in the opposite direction of the other side since the magnetic field vectors are in opposite directions?

Depend on your viewpoint.  The field passes through the ring magnet so it is in the same direction both sides.  But if you look at it as a field coming out of the ring then in one side the vector points outward and on the other side it points inward and I suppose you could consider that as "opposite".   It is easy to show that the new induction term yields the same polarity from brush to brush over top and bottom halves of the slip ring, so it induces voltage into the external circuit.  Of course that comes from the forces on the mobile conduction electrons.  On the lattice ions there is of course forces in the opposite directions so the reaction on the slip ring is a sideways one trying to move it towards one of the brushes.  And thereby comes the OU, there is no reaction torque, no circular force around the slip ring.  Luckily it can be shown that when the generator is driving current through a load electrons hop from brush to slip ring or vice versa and therefore undergo acceleration, and the radiated E field from that acceleration acts on the atomic electron circulations in the magnet extracting energy that exactly accounts for the generator output.

QuoteThe effect is the same, either way, but how do you determine which field is producing the induction in the ring?

There is only one field there, the magnetic vector potential A field.  There is no E field  and there is no B field from the magnet

QuoteWhat if you shielded the magnet from electric fields?  Would the MVP still be there?

The A field can't be shielded like that, so it would have no effect.
Smudge

Phipps noted that there is a back EMF associated with the Marinov Generator and provides a rough formula for it in the summary of this paper.

Quote from: Grumpy on 2018.05.03, 16:38:55
Sorry for the double post, but is the magnetic vector potential considered a "force"?

It's units are the same as momentum per unit charge.

It is considered to be the source of so called "hidden" momentum, or "canonical" momentum, a form of momentum that is not related to velocity or mass.  Hence a particle with charge q within an A field has this hidden momentum of value qA  pointing along the A direction.  The particle obeys force=rate-of-change of total momentum, F=d(mv+qA)/dt (bold characters are vectors).  So yes, force is involved.

You already know that with transformer induction the secondary coil is outside the core, is not within a B field, and the alternating A field is what drives the electrons in that coil.  There you have a static coil and a time changing A field producing the force.  In the Marinov generator the A field is static but it is non-linear, so an electron moving through it "sees" a time changing A value because of that non-linearity, hence it gets a force.  It is that form of induction that is not recognized classically, although it is the subject of much debate.  I find it odd that the debate has not been settled by experiments such as those suggested here.  It is only the Marinov motor that has been played with and that has given odd results, just ask TK!  There the electron velocities are very low and there is an assumption (wrongly IMO) that the tiny longitudinal induction will cause the slip ring to rotate.   Just how the force translates to the lattice ions is beyond me.
Smudge

Quote from: Grumpy on 2018.05.03, 19:09:44
Phipps noted that there is a back EMF associated with the Marinov Generator and provides a rough formula for it in the summary of this paper.

That was in 1998 but AFAIK the independent observations have never been published.  That formula is classically what you get for any efficient generator where output watts equate to shaft input power, and IMO Phipps used that to hint that the observations were on a generator that simply obeyed COE.  I would be most interested to see someone else's work on the Marinov generator, but years of searching have not revealed anything.
Smudge

I don't know if this paper is helpful, as I just found it and have not read it. 

https://skemman.is/bitstream/1946/29521/3/KOK_MagneticVectorPotential.pdf

Hal Puthoff is on page 47 (communication system).

See bottom of page 64:

QuoteIn the authors view, the most significant part of the work is the factorization of the convective derivative of the magnetic vector potential resulting in the force terms in Equation (5.11a) to (5.11d). These could in theory lead to new techniques of current induction without magnetic fields, by the movement of particles within the magnetic vector potential field, possibly practical in motor and generator design

Quote from: Grumpy on 2018.05.03, 23:09:15
I don't know if this paper is helpful, as I just found it and have not read it. 

https://skemman.is/bitstream/1946/29521/3/KOK_MagneticVectorPotential.pdf

Hal Puthoff is on page 47 (communication system).

See bottom of page 64:

Note the date, 2018.  So this is quite new.  Another extract from the document is

QuoteThe truthfulness of the above statements will have to be verified with experiment.
New ways of induction in the absence of magnetic fields would open doors in generator
and motor designs. This matter is thus quite significant. The expansion of the
convective term for the magnetic vector potential into (5.11a) to (5.11d) is original to
the best of the authors knowledge.

Well that expansion is not new, I have several papers from other authors where that expansion is followed (including Phipps and written 20 years ago!).  IMO the jury is still out as to whether the Marinov generator is feasible and I am hoping for evidence that it is.  If I am right it could lead to OU but there would be engineering challenges.  A machine built by the likes of the researchers here on this forum would likely not be OU because of those challenges, brush friction and voltage drop would degrade the performance to below 100% efficiency.  But it appears to me that the device is inherently OU and those degradations could be overcome.

Thanks G for bringing that paper to my attention, I'll add it to my stack of papers on the subject.
Smudge

P.S.  I have met Hal Puthoff and corresponded with him.

Perhaps there is a way to manipulate space and the magnetic vector potential that occupies it rather than trying to utilize the feeble manifestation of the magnetic vector potential in  the Marinov Generator.

My point with this statement is that the output of the Marinov Generator is so low. 

Quote from: Grumpy on 2018.05.03, 23:09:15
I don't know if this paper is helpful, as I just found it and have not read it. 

https://skemman.is/bitstream/1946/29521/3/KOK_MagneticVectorPotential.pdf

Hal Puthoff is on page 47 (communication system).

See bottom of page 64:

Thank you as well Grumpy. Zimmerman I have had dealings with on Researchgate, he writes patents like they are going out of fashion. As you know I am working on the magnetic vector of a loop, as in transmitting loops and wave interference as my idea on the TPU, it is looking good ATM, but still a ways to go me thinks ;)

Regards

Mike 8)

from the paper I linked above, page 51:

Quote
Interestingly there exists a controversial generator/motor design known as the Marinov motor or Siberian Coliu which was explained with the force term
∇(qv·A) by Stefan Marinov. Theoretical physicist James P. Wesley (1998) agreed with the analysis and noted that the Hopper-Monstein effect, another example of induction with a zero magnetic field, could be explained by it as well.

I don't recall what the Hooper-Monstein effect is.  I'll have to look that up.

http://jnaudin.free.fr/html/hoopmnst.htm

Regards

Mike 8)

I looked up the Hooper-Monstein Effect and find it very interesting.

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0ahUKEwjgqPGvuPTaAhWKiOAKHYXiBhAQFggwMAE&url=http%3A%2F%2Fsciencepublishinggroup.com%2Fbook%2Fdownload%3FchapterId%3D4953%26stateId%3D8000%26fileType%3D4&usg=AOvVaw3SfvZsglGQkUyvnqXndny1

See image page 171, quote below on page 172:

QuoteThe experiment shows in this case that – in reality, the electric potential and
the electric current are induced by the magnetic potential A and not by the
magnetic induction B=rot A, the A vector being circulary oriented around the
magnet axis and of mutually opposed circular sense for the two magnets, the
total magnetic potential in the P-point being of value: AT =A1+A2=2A, also
when the magnets are rotated in the same sense, so-the electro-magnetic
induced effect in the P-point is of double value, in this case, [3].

I like Tinman's idea for using a plastic food bowl for a quick budget experiment.  His idea of using two horseshoe magnets should work OK but of course the output is low voltage DC.  I think an AC experiment is worthwhile because if this really works and is OU then it will need AC generation.  This is easily achieved by using a soft ferromagnetic ring core driven with AC.  Many of you will have small ring cores that would fit into a plastic drinking glass, so that could be a route to a budget experiment, see images below.  And you will have oscilloscopes that can easily measure low voltage AC to compare with the input.  If it works there should be low voltage AC output that is independent of frequency, in phase with the input current and proportional to rotation speed.  If that is achieved it shows there to be an unusual form of induction that could be OU in that output energy is not derived from the input to the coil, but comes from whatever space energy keeps electrons perpetually spinning and orbiting.  Anyone up for this?
Smudge

Bump!

This thread seems to have died for some reason.  As the final post was for an interesting experiment that would produce low voltage AC (that can be transformed), and not low voltage DC (that cannot be transformed), I am surprised that no one has taken up the offer to make one.  Alas I can't do it myself as family circumstances have me living with our granddaughter and I have no tools or equipment.

Smudge

Hi Smudge,

I always thought this was an interesting experiment to replicate. Marinov was certifiable, but he occasionally had a good idea..

F.

Quote from: orthofield on 2019.04.25, 15:38:21
Hi Smudge,

I always thought this was an interesting experiment to replicate. Marinov was certifiable, but he occasionally had a good idea..

F.

Hi Orthofield

The subtlety of Marinov's motor is that the conductors do not bathe in the magnetic field and yet, a force is exerted, which can no longer be defined as Lorentz's force.
However, the weakness of the effect leads me to believe that it is of the second order, or non-existent if it is only an experimental artifact as claimed by some physicists who have refuted this motor.
In any case I think that we should be able to find a first-order effect, i. e. where the force depends on B, but always with B=0 in the volume of the conductors, in the same way as B=0 at the secondary winding of a transformer, the field being confined in the magnetic core.
In this case, the natural explanation would be the vector potential A. At my knowledge, there is no motor using A instead of B, while it is very common with induction. This would be a novelty to use A for a mechanical effect.

Quote from: F6FLT on 2019.04.29, 12:46:13
In this case, the natural explanation would be the vector potential A. At my knowledge, there is no motor using A instead of B, while it is very common with induction. This would be a novelty to use A for a mechanical effect.

The Aharanov-Bohm effect is well known and there are several authors who show that this can be explained by change of electron velocity, i.e. a mechanical effect.  In my view of the Marinov motor the longitudinal effect on the moving electrons should show up as a change in resistance, not movement of the whole lattice to include the ions.

What interests me is the Marinov generator where the lattice of ions plus conduction electrons is driven at speeds far in excess of drift velocity.  Now the purported effect coming from movement through a non-uniform A field will apply oppositely directed forces to ions and electrons thus causing the potential difference across the brushes.  My efforts in this field are reported in this thread where I obtained definite results, but using magnets that also supplied a B field so they are not conclusive.  What I am hoping for is an experiment using a ring core where there is no external B field.

Smudge

The Aharanov-Bohm effect is slight. I am looking for an effect as powerful in mechanics as that of the potential vector A on the electrons of a coil traversed by a variable magnetic field, producing a current while B=0 in the volume of the wire.
The same force that generates the current should be obtained transversely, resulting in a mechanical force on the conductor. It is the idea of implementation that I am looking for on my side in terms of a new generation of homopolar generator.
The only thing in common with Marinov is that the conductor on which the mechanical force is exerted is not immersed in a magnetic field. Marinov opened the way with his idea, but his idea uses the spatial gradient of A, and I think we can use dA/dt as in induction, which should make possible a much higher mechanical force.

Hi Smudge,

In the case of Marinov, the field lines of A must be difficult to define mathematically. Even with a magnetic field confined in a circular torus, the field lines of A must be something like ellipses, and if we have to make elliptical integrals, we are not out of the woods...

So to move forward step by step, and to verify that the potential vector approach is possible, I am trying to describe the Faraday disc in terms of potential vector, without using the magnetic field.
This is surely possible because the circular symmetry of A allows us easy integrals.

Indeed, if we can describe Faraday's disc without using the magnetic field, we obtain the proof of concept of a motor/generator with B=0 and A≠0, and the door is also open for many new innovative homopolar systems in addition to Marinov's.

Have you already done it for Faraday's disk, or can you try to do it? That's what I'm trying to do, but I'm not doing it yet.

Hi F6,

I have looked into using the convective derivative of the vector potential A field to describe the Faraday disc induction but came to an abrupt halt part way through because it predicts half the voltage obtained from the usual v X B induction.  I am posting the unfinished document here.  I also post another unfinished document on the Marinov generator where the item of interest is the appendix which shows that using (v.del)A=del_A(v.A)-v X B has some hidden subtleties that may explain the discrepancy.  You will see from my ramblings that I don't fully understand differentiation of vectors.
Smudge

Quote from: F6FLT on 2019.05.05, 11:30:48
Hi Smudge,
In the case of Marinov, the field lines of A must be difficult to define mathematically.

They can be derived quite easily using finite element programs.  I use the free FEMM for simulations which admittedly is only a 2D model but can yield useful results.  Because of the math similarity governing H field lines around a conductor carrying current and A field lines around a core carrying flux you can use the former to predict the latter.  The A field lines in my paper "On Electrodynamic Formula" were obtained from FEMM that way, I simply used the H field plots around currents, where 1 amp represents 1 Weber and the resulting amps/meter are read as Weber/meter.   FEMM also does the integration for you along selected contours.

Smudge

Quote from: Smudge on 2019.05.05, 14:45:23
Hi F6,

I have looked into using the convective derivative of the vector potential A field to describe the Faraday disc induction but came to an abrupt halt part way through because it predicts half the voltage obtained from the usual v X B induction.
...

Hi Smudge,

Thank you for the documents, excellent work.
For Faraday's disc, I get the same result and I don't see why we have this 1/2 factor in the solution with A.
We absolutely have to solve this. If we don't have the correct solution by A in the simple case of Faraday's disc, we have missed something big, and we risk failing in the more complex problems.

I wonder if the spatial gradient of A should be explicitly used. But I'm progressing very slowly and you seem to be well ahead of me, so go on with your idea.

Concerning the calculation of elliptical integrals, I meant to find the analytical solution. Of course we can do the numerical calculation by software.

Having given more thought to the discrepancy when using the convective derivative of the A field to calculate voltage from the Faraday disc homopolar generator I have arrived at a new perspective.  See my new paper here for my definition of the E field.
Smudge

Quote from: Smudge on 2019.05.14, 19:08:58
Having given more thought to the discrepancy when using the convective derivative of the A field to calculate voltage from the Faraday disc homopolar generator I have arrived at a new perspective.  See my new paper here for my definition of the E field.
Smudge

Hi Smudge,

I said, "I wonder if the spatial gradient of A should be explicitly used," and here's why.
In your work, you write "plus another term that takes into account the evolution of A as seen by an electron moving through a non-uniform A field". I think there is a weakness here because the analysis must be done in only one referential frame, usually the observer's.

When using E=-∂A/∂t, it is the time variation seen by the observer that sees the force F=q.E on the charge. A and E are related to the observer, not the charge. The charge does not see the same A because it is mobile, and worse, because it is accelerated, i.e. its reference frame is not even inertial.
The spatial gradient of A, if it is to be introduced, must be that seen by the observer, not the charge which can't see a spatial gradient because it is assumed to be point, it can only see a temporal gradient, the electric field. If a spatial gradient gave rise to an electric field, then a charge placed far from an infinite (very long) solenoid would be attracted to it, but this is not what we observe.

I think the problem turns more on how to integrate on a loop when the circuit of this loop has parts that move relative to each other. We cannot assume that the electrons will just follow the radius of the disc between the sliding contacts. The electrons are forced to move tangentially because of the rotation of the disc, so they will follow a non-radial path statistically longer than the radius. I admit I don't know how to translate that into the equations.

Quote from: F6FLT on 2019.05.19, 10:47:21
Hi Smudge,

I said, "I wonder if the spatial gradient of A should be explicitly used," and here's why.
In your work, you write "plus another term that takes into account the evolution of A as seen by an electron moving through a non-uniform A field". I think there is a weakness here because the analysis must be done in only one referential frame, usually the observer's.

And if the observer is moving through a non-uniform A field he will see a temporal gradient as his local A field changes value.  Or are you suggesting that the moving observer will not see the same A field, his A field will be different?  I could agree with you if the source of the A field, which is itself some moving charge, is moving at low velocity and you invoke relativity, i.e your observer's movement alters the relative motion between source and observer.  But our A fields come from spinning or orbiting electrons, and their velocities are so tremendous that our puny movement is immaterial.  That is why in the Faraday homopolar machine it matters not whether the magnet is fixed or moves with the disc.  Spinning the magnet at our puny RPM's does nothing to the relative motion between source and observer.

QuoteIf a spatial gradient gave rise to an electric field, then a charge placed far from an infinite (very long) solenoid would be attracted to it, but this is not what we observe.

Why would the charge be attracted to the solenoid?  The A field forms circles around the solenoid axis, and for the infinitely long solenoid that field is curl free, there is no B field there.  Yes the circular A field increases in magnitude as you move closer radially, but that seen as a temporal increase is a transverse force on the moving electron, and that is not an attractive force.

QuoteI think the problem turns more on how to integrate on a loop when the circuit of this loop has parts that move relative to each other. We cannot assume that the electrons will just follow the radius of the disc between the sliding contacts. The electrons are forced to move tangentially because of the rotation of the disc, so they will follow a non-radial path statistically longer than the radius. I admit I don't know how to translate that into the equations.

I think that spiralling action has led people to believe that the homopolar disc machines are OU.  My Marinov generator does not have that problem, and unlike other claims for OU I can show where the energy comes from.  I can show the possibility for the anomalous energy to be obtained from the PM's spinning or orbiting electrons responsible for its magnetization, the machine tries to slow down those spins.

Smudge

Quote from: Smudge on 2019.05.19, 16:27:59
And if the observer is moving through a non-uniform A field he will see a temporal gradient as his local A field changes value.  Or are you suggesting that the moving observer will not see the same A field, his A field will be different?    I could agree with you if the source of the A field, which is itself some moving charge, is moving at low velocity and you invoke relativity, i.e your observer's movement alters the relative motion between source and observer.  But our A fields come from spinning or orbiting electrons, and their velocities are so tremendous that our puny movement is immaterial... 

The question of the origin of A is not relevant, because the potential vector is, like a field, a local property of space at the position of the charge. If the charge is one light year from the source of A, and the source of A is anihilated, the charge will continue to see A for one year, and if it starts moving in A, it will no longer see the same A. So it doesn't matter which source of A. In relativity A is a 4-vector to which the Lorentz transforms apply, and the change is quite simple by matrix calculation:

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

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.

Quote
Why would the charge be attracted to the solenoid?  The A field forms circles around the solenoid axis, and for the infinitely long solenoid that field is curl free, there is no B field there.  Yes the circular A field increases in magnitude as you move closer radially, but that seen as a temporal increase is a transverse force on the moving electron, and that is not an attractive force.

Around the solenoid, the equipotentials of A are concentric circles that weaken proportionally with distance.
If the spatial gradient has an effect on the charge, it is because dA/dx results in an electric field. In our particular case, the differential can only be made on an axis transverse to A, i. e. x is radial. Unlike t, x is not a simple scalar, it has a direction. I don't see what makes you think that the E field related to dA/dx should be along A.

Quote
I think that spiralling action has led people to believe that the homopolar disc machines are OU.  My Marinov generator does not have that problem, and unlike other claims for OU I can show where the energy comes from.  I can show the possibility for the anomalous energy to be obtained from the PM's spinning or orbiting electrons responsible for its magnetization, the machine tries to slow down those spins.

Smudge

I'm not that enthusiastic. I note that the effect behind Marinov's device is much weaker, at equal field strength, than Faraday's. So either we have an artifact due to a residual field, or the effect is real but not related to the first power of the spatial gradient.

However, the Marinov's idea raises a much more interesting question, to which a positive answer opens up much more prospects than the Marinov generator alone: can we create a homopolar generator with mobile conductors through a non-zero vector potential but a zero magnetic field B?
I am preparing a schematic of an experiment to prove it, it will be ready today, maybe even this morning.

Quote from: F6FLT on 2019.05.20, 08:24:49
The question of the origin of A is not relevant, because the potential vector is, like a field, a local property of space at the position of the charge. If the charge is one light year from the source of A, and the source of A is anihilated, the charge will continue to see A for one year

I fail to see how this affects the argument.  We are dealing with systems where the distance between the source and the charge is so small that the propagation time is negligible.

Quoteand if it starts moving in A, it will no longer see the same A.

Explain why it will no longer see the same A.  You quote the Lorentz transform which invokes beta and gamma but since we are dealing with electron velocities that are small compared to c, beta tends to zero and gamma tends to unity.  So IMO we can forget the Lorentz transform, it has negligible effect and to all intents and purposes the moving electron sees the unmodified A.

Quote
Around the solenoid, the equipotentials of A are concentric circles that weaken proportionally with distance.
If the spatial gradient has an effect on the charge, it is because dA/dx results in an electric field. In our particular case, the differential can only be made on an axis transverse to A, i. e. x is radial. Unlike t, x is not a simple scalar, it has a direction. I don't see what makes you think that the E field related to dA/dx should be along A.

Inside the solenoid the equipotentials of A are also concentric circles but they strengthen proportionally with distance.  This also applies to the A field in the Faraday disc.  As you well know the force on any radially moving electron there is not radial, it is always transverse to the movement.

QuoteI don't see what makes you think that the E field related to dA/dx should be along A.

For E=-dA/dt the E is along A.  If you examine the components of (v.del)A the E is along A.

QuoteI note that the effect behind Marinov's device is much weaker, at equal field strength, than Faraday's.

Not true.  What equal field strength are you using?  The Marinov generator discussed in this thread has zero B field at the moving electrons.

Quote
However, the Marinov's idea raises a much more interesting question, to which a positive answer opens up much more prospects than the Marinov generator alone: can we create a homopolar generator with mobile conductors through a non-zero vector potential but a zero magnetic field B?

IMO the Marinov generator is just that.

QuoteI am preparing a schematic of an experiment to prove it, it will be ready today, maybe even this morning.

I look forward to seeing it.

Smudge

Quote from: Smudge on 2019.05.20, 10:51:23
I fail to see how this affects the argument.  We are dealing with systems where the distance between the source and the charge is so small that the propagation time is negligible.

This argument was just related to your comment: "But our A fields come from spinning or orbiting electrons". A potential as well as a field is a tool to avoid to discuss the question of the source and of the propagation of an effect from a source.
If we deal with the potential vector, the question of the source is off topic.

Quote
Explain why it will no longer see the same A.  You quote the Lorentz transform which invokes beta and gamma but since we are dealing with electron velocities that are small compared to c, beta tends to zero and gamma tends to unity.  So IMO we can forget the Lorentz transform, it has negligible effect and to all intents and purposes the moving electron sees the unmodified A.

Electromagnetism effects are relativistic effects. It is a common mistake to think that the low electron velocity could be negligible while it is the only cause of the magnetic field. I found the best explanation is the book "Claustrophobic physics (http://exvacuo.free.fr/div/Sciences/Cours/R.Paul%20Bickerstaff%20-%20Claustrophobic%20physics.pdf)" from Paul Bickerstaff.
As the question comes up regularly, I have attached an extract from his chapter on currents, with the question of the force between 2 parallel wires carrying a current.

I quote him:
"So the magnetic field experienced by the test charge is correctly understood as arising from an electric force in another frame due to a line of net charge density γ.β².λ0 [...]. The relative strength of the magnetic and electric force is thus of the order of β² where β is determined by the drift velocity of the electrons. As remarked above, this is of the order of just a few mm/s and so
        β ≈ 10^-11
for typical current bearing wires. The magnetic force they exert is therefore just 10^-22 times that of an electric force due to just the positive (or negative) charges. That the magnetic force is observed at all (and is appreciable) is due to the enormous amount of charge carried by a number of electrons of the order of the Avogadro's number".

In other words, the electrical force is really a huge force and the large number of charges makes visible the smallest relativistic effect. Without the natural balance of positive and negative charges in nature, this force would prevail everywhere.
The electrical effects come from the separation of charges.
Magnetic effects come from relative speeds between charges, the relativistic effects cancelling the apparent balance of the charges, in fact the equilibrium of their electric field.

Quote
Inside the solenoid the equipotentials of A are also concentric circles but they strengthen proportionally with distance.  This also applies to the A field in the Faraday disc.  As you well know the force on any radially moving electron there is not radial, it is always transverse to the movement.For E=-dA/dt the E is along A.  If you examine the components of (v.del)A the E is along A.

My interrogation was general. I wrote dA/dx by simplification but a spatial gradient is dA/dl where A and l are vectors. And we know there is no field gradient along A, so the question of the gradient direction arises, and this question doesn't depend on the speed of the charge that moves in.
A and dA/dt are independent of the load, they are evaluated by the observer.
So the spatial gradient of A should be as well. Let's not forget that the electric field is a property related to the place. If we assume that the spatial gradient gives rise to an electric field, this field cannot depend on the test charge that is placed in it. If charges at different speeds see different fields, this is understandable, it is related to relativity, but in any case the observer should only see a single electric field!

The inconsistency is due to the fact that I have already mentionned, that of calculating the spatial gradient in the referential of the charge instead of that of the observer, and mixing it with dA/dt or VxB which are relative to the observer's referential.
Can you give the spatial gradient of A, viewed in the referential frame of the observer ?

Quote
IMO the Marinov generator is just that.I look forward to seeing it.

Smudge

The Marinov generator is just that but perhaps there is simpler. In the Marinov generator there is a spatial A gradient, not a constant A. In the Faraday disk, A is constant, spatially and temporally. Just keep A and remove B in the Faraday disk, and you have a possible homopolar generator working on A, and not on a gradient of A, which is different from Marinov. I don't know yet the result. See my next post.

Here is the simplest idea of using A instead of B for a homopolar generator.

Take a infinite (long) cylinder magnet (like an infinite solenoid carrying a constant current). Make a hole in the disk and put the magnet through it. Place the sliding contacts between the inner and outer diameter, and rotate the disk, or the disk with the magnet as on the picture.

As for an infinite solenoid, there is no B field outside the magnet, but at the position of the disk, the vector potential is not null. If we have a voltage, it coudn't have the Lorentz force as explanation, unlike the Faraday disk. It could be only the result of the rotation in the vector potential.

But will we have this voltage?

If the Faraday disc can be explained with only the vector potential, then there are good reasons for this configuration to work, because unlike the magnetic field, there is continuity of A when one passes from the inside of the magnet to the outside.


(To avoid experimental biases as a magnetic field leakage, the long magnet could be replace by a toroid magnet where the B field would be confined).

See the attached document:

(Sorry if I posted this before, I have a poor memory.)

Quote from: Grumpy on 2019.05.20, 15:08:06
See the attached document:

(Sorry if I posted this before, I have a poor memory.)

Grumpy,
You didn't post it but I did in the first post of this thread.  That's my paper. :)
Smudge
Edit.  Yes you did post it so that's three postings within this thread.

Quote from: F6FLT on 2019.05.20, 14:23:56
Here is the simplest idea of using A instead of B for a homopolar generator.

Take a infinite (long) cylinder magnet (like an infinite solenoid carrying a constant current). Make a hole in the disk and put the magnet through it. Place the sliding contacts between the inner and outer diameter, and rotate the disk, or the disk with the magnet as on the picture.

As for an infinite solenoid, there is no B field outside the magnet, but at the position of the disk, the vector potential is not null. If we have a voltage, it coudn't have the Lorentz force as explanation, unlike the Faraday disk. It could be only the result of the rotation in the vector potential.

But will we have this voltage?

If the Faraday disc can be explained with only the vector potential, then there are good reasons for this configuration to work, because unlike the magnetic field, there is continuity of A when one passes from the inside of the magnet to the outside.


(To avoid experimental biases as a magnetic field leakage, the long magnet could be replace by a toroid magnet where the B field would be confined).

If anyone does this experiment I would be interested in the result.  But even if it yields zero voltage there is still the possibility that a longitudinal force (along the velocity direction) could be present in an appropriately shaped A field.  The trick of using vector identities but with some parts removed (like the velocity derivatives) seems a poor way of describing the physics.  But if you are going to do this why not remove the transverse components and leave the longitudinal ones.  You need a different experiment to show up that longitudinal force, an experiment like the Marinov generator. :)

Smudge

Quote from: F6FLT on 2019.05.20, 14:21:54
In the Faraday disk, A is constant, spatially and temporally.

It is not constant spatially.  If it were there would be zero B field.  But I guess you know this and didn't mean to write that.

Smudge

Quote from: Smudge on 2019.05.20, 18:45:05
It is not constant spatially.  If it were there would be zero B field.  But I guess you know this and didn't mean to write that.

Smudge

I haven't been explicit enough. Of course A is not constant everywhere in space. I just want to say that unlike Marinov's device, the conductor movement is done in a constant A: each point of the disc rotates in a constant potential because A is circular and concentric to the axis of rotation.
Naturally each point is not in the same A, it depends on its position along the radius, but the mechanical rotational movement keeps the electron in the same A, so a spatial gradient which would be related to the rotation can't exist and be the cause for a current. In other words, the electron is not forced in a vector potential gradient.

Quote from: F6FLT on 2019.05.20, 19:12:09
I haven't been explicit enough. Of course A is not constant everywhere in space. I just want to say that unlike Marinov's device, the conductor movement is done in a constant A: each point of the disc rotates in a constant potential because A is circular and concentric to the axis of rotation.
Naturally each point is not in the same A, it depends on its position along the radius, but the mechanical rotational movement keeps the electron in the same A, so a spatial gradient which would be related to the rotation can't exist and be the cause for a current. In other words, the electron is not forced in a vector potential gradient.

The same argument applies to the force between two infinitely long parallel current carrying conductors, Ampere's force law.  The electrons are moving along a constant A field line, yet they endure a force at right angles to that movement.  When you examine the individual components of the flux cutting v X B force as expressed by movement through the A field you find ones like E_x=v_y*dA_y/dx that brings in the spatial gradient at right angles to the movement.  So even though the electron is not moving through that spatial gradient its presence influences the force on the electron.  As you pointed out, the electron as a point particle can't "see" or "Know" that spatial gradient exists, it really is a matter of relativity and different frames of reference, but nevertheless we do have tried and tested vector-math formula to predict those forces and they do use that spatial gradient. 

Your suggested experiment has the same problem, the disc rotation moves the conduction electrons along constant A field lines.  If that really does yield an output voltage that can drive current through a load, the question must be asked "where does the energy come from".   IMO it comes from the electron spins or orbits responsible for the magnetic field in the core, as I have repeatedly tried to say in various ways.  Electrons passing from a brush contact onto a moving conductor must endure an acceleration, and therefore "radiate" an electric field.  It is the electric near-field that interests me and I think it possible that it has unusual characteristics in that a small closed circuit near the brush contact will obtain an induced unidirectional voltage impulse for each electron that jumps across.  For a continual stream of electrons that becomes a DC voltage.  Note that there is no dB/dt present to invoke the usual transformer induction rule for closed circuits.  Each magnetic dipole atomic electron circulation within the core gets this induced voltage attempting to slow down the spin or orbit, which it can't do.  So these spins are truly miniature quantum dynamos.  I attach an old paper written some years ago where I try to explain all this.   Also another old paper on possible experiments to discover longitudinal induction where I see a did a crude experiment using a square NdFeB magnet.

Smudge

Quote from: Smudge on 2019.05.21, 11:04:05
The same argument applies to the force between two infinitely long parallel current carrying conductors, Ampere's force law.  The electrons are moving along a constant A field line, yet they endure a force at right angles to that movement.  When you examine the individual components of the flux cutting v X B force as expressed by movement through the A field you find ones like E_x=v_y*dA_y/dx that brings in the spatial gradient at right angles to the movement.  So even though the electron is not moving through that spatial gradient its presence influences the force on the electron.  As you pointed out, the electron as a point particle can't "see" or "Know" that spatial gradient exists, it really is a matter of relativity and different frames of reference, but nevertheless we do have tried and tested vector-math formula to predict those forces and they do use that spatial gradient. 

Apart from a mechanical force, there is only one force capable of moving an electron, the force F=q.E.
If we have a current, we have a force, so we have an electric field E seen by the electron.
In the case of the classical Faraday disc, this electric field E seen from the charge is VxB where v and B are seen by the observer.
In the case of the Faraday disc with an out-of-field conductor, as in my experimental proposal, VxB can no longer be used.
So I agree with you. When dealing with A, your idea that E is due to the transverse gradient of A, in this case as in the other, is a credible hypothesis to be tested, with real chances to be verified.

I won't be able to do this experiment because I'm too bad at mechanics. The long magnet (to be made from a series of smaller ones) will be difficult to assemble and rotate quickly enough. There is the problem of the magnet end leakage field which can create artifacts. And using a torus instead of the long magnet drastically complicates the rotation, because you need a fixed toroidal magnet, you lose the axis of rotation, so you need a transmission belt to rotate the disc. I don't feel like I can do it.

Quote
Your suggested experiment has the same problem, the disc rotation moves the conduction electrons along constant A field lines.  If that really does yield an output voltage that can drive current through a load, the question must be asked "where does the energy come from".   IMO it comes from the electron spins or orbits responsible for the magnetic field in the core, as I have repeatedly tried to say in various ways.  Electrons passing from a brush contact onto a moving conductor must endure an acceleration, and therefore "radiate" an electric field.  It is the electric near-field that interests me and I think it possible that it has unusual characteristics in that a small closed circuit near the brush contact will obtain an induced unidirectional voltage impulse for each electron that jumps across.  For a continual stream of electrons that becomes a DC voltage.  Note that there is no dB/dt present to invoke the usual transformer induction rule for closed circuits.  Each magnetic dipole atomic electron circulation within the core gets this induced voltage attempting to slow down the spin or orbit, which it can't do.  So these spins are truly miniature quantum dynamos.  I attach an old paper written some years ago where I try to explain all this.   Also another old paper on possible experiments to discover longitudinal induction where I see a did a crude experiment using a square NdFeB magnet.

Smudge

I don't agree. We are only dealing with a relativistic effect between two electron flows of different speeds. A force is exerted between the two parts of the circuit, and nothing acts on the source of the field or the potential. As for the normal Faraday disk, the energy comes only from the mechanical energy to rotate the disk. Even if the disc no longer rotates in a magnetic field, the current still generates the same magnetic field in both cases, so it will oppose rotation as in the ordinary Faraday disk. The magnetic dipoles including spins are just the vectors of the change of the mechanical energy into electrical energy.

We should think of simpler ways than mechanics to highlight all this, for example with a solid state device that would make electrical induction from the spatial gradient of A.

Quote from: F6FLT on 2019.05.21, 13:25:43
I don't agree. We are only dealing with a relativistic effect between two electron flows of different speeds. A force is exerted between the two parts of the circuit, and nothing acts on the source of the field or the potential. As for the normal Faraday disk, the energy comes only from the mechanical energy to rotate the disk. Even if the disc no longer rotates in a magnetic field, the current still generates the same magnetic field in both cases, so it will oppose rotation as in the ordinary Faraday disk.

Then we will have to agree to disagree. You say a force is exerted between the two parts of the circuit, and that force is on the moving electrons.  That is the moving electrons in one part and the moving electrons in the other part.  It so happens that Nature in her perverse way has arranged things so that in our usual machines such as the Faraday disc when we extract electrical energy from one part the force acting on the electrons in the other part creates mechanical drag so we have to supply mechanical force and that is part of the energy balance.  However there are other machines where the force on the "driving" electrons is along the conductor even if it is circular (a coil) and we have to supply electrical energy to complete the energy balance.

QuoteThe magnetic dipoles including spins are just the vectors of the change of the mechanical energy into electrical energy.

That is a very limited view of magnetic dipoles.  To me a magnetic dipole is a current loop and like any current loop it can get an induced voltage that either aids or opposes the current.  Hence whatever current generator is there driving current around that loop, it can either supply energy to whatever is creating that induction or receive energy from it.  I can argue that atomic dipoles are no different and that in many cases they already do just that, supply energy or receive energy, but that energy exchange is hidden from view because it doesn't appear in the usual energy audit, over complete cycles the exchange sums to zero, i.e they take back as much energy as they give out.  Note this is not any form of dipole rotation or flipping, this is not mechanical energy related.  The fact that this is happening under our very noses, and we ignore it, is a big mistake on our part.  We should be looking for ways to break that reciprocity, and I think that can happen when we have atomic dipoles close to brush contacts that are carrying DC.

QuoteWe should think of simpler ways than mechanics to highlight all this, for example with a solid state device that would make electrical induction from the spatial gradient of A.

Well my suggestion for using a change of drift velocity at a junction between a normal conductor and a super-conductor is one possibility.  I think there are others, for instance the drift velocity along a conductor surface where the conductor is charged highly negative will be much greater than the normal drift velocity.  That is simply because there you have a region with electrons just above the rough surface (rough at atomic dimension scale) hence the collision rate with ions is much reduced.  It must be possible to use this greater electron speed at a junction with an uncharged conductor to get acceleration there, and use that instead of a junction of a brush against a moving conductor.

Smudge 

Can you use a simple experiment like this one?

https://pdfs.semanticscholar.org/88a9/37981ae9f9663c591f9aca5db108faf122f0.pdf

see page 203 of the linked paper:

QuoteThe writer is not the first observer in this country to confirm the existence of nonzero torque in the Marinov motor. As reported by Kooistra (13),  Tom  Ligon  was  the  first  to  recognize  that  an  easy way to test the concept is to suspend a toroidal magnet "armature" shaped somewhat as in Fig. 1 of Wesley's  report (1),  within  a  horizontal  copper  ring,  fixed  in  the  lab,  to  which  current  leads  are  attached. This eliminates the need for brushes, and emulates one very practical brushless motor design (which,  however,  requires  periodic  DC  polarity  reversals).  Ligon  observed  vigorous  turning  of  the armature up to almost 90 degrees when strong DC was put through the ring. At 90 degrees there is a torque null, and beyond that a counter-torque.

Quote from: Smudge on 2019.05.21, 14:51:51
Then we will have to agree to disagree. You say a force is exerted between the two parts of the circuit, and that force is on the moving electrons.  That is the moving electrons in one part and the moving electrons in the other part...

That's what I'm saying, I confirm. You forgot VxB: electrical current and the mechanical force are transversal.
If you pass current through a Faraday disc, it turns into a motor, so the current in the circuit creates a torque.
When you use the disc as a generator, and draw current from a load, this current acts exactly the same way as in the Faraday disc as a motor, and the direction of the currents causes the torque to oppose rotation.
All this is completely reciprocal.

Quote
...To me a magnetic dipole is a current loop and like any current loop it can get an induced voltage that either aids or opposes the current.

Magnetic dipoles are current loops. But why do you thing that there would be induced voltage in these dipoles in the case of the Faraday disk? Why do you think that induced voltage would have an effect on the electron spins?...

The magnet may or may not rotate because there is no interaction of any "counter"-field with those of the magnet's dipoles. This is due to the fact that the magnetic force F=VxB doesn't work, the force being perpendicular to the displacement, so the reaction is not here.
Otherwise rotating a cylindrical magnet in front of another that is coaxial would cause a torque from one to the other. The experiment has been done many times, there is no effect. I'm afraid you're going to go into highly speculative speculation   :).

Replace the permanent magnet of a Faraday disk by a solenoid. I'm sure you will see no induced voltage.

Will the Marinov device work if the conductive ring encircles only one side of the toroidal magnet?

Quote from: Grumpy on 2019.05.21, 18:48:58
Will the Marinov device work if the conductive ring encircles only one side of the toroidal magnet?

It won't, unless the sliding contacts are between the inner and outer edge of the rotating ring, and the ring is sufficiently thick. This is what I proposed here (https://www.overunityresearch.com/index.php?topic=2470.msg74662#msg74662), simply replace the long magnet by one of the two sections of the Marinov's magnetic circuit.

Quote from: F6FLT on 2019.05.21, 15:58:38
That's what I'm saying, I confirm. You forgot VxB: electrical current and the mechanical force are transversal.
If you pass current through a Faraday disc, it turns into a motor, so the current in the circuit creates a torque.
When you use the disc as a generator, and draw current from a load, this current acts exactly the same way as in the Faraday disc as a motor, and the direction of the currents causes the torque to oppose rotation.
All this is completely reciprocal.

I didn't forget and I agree with everything you say here. 

QuoteMagnetic dipoles are current loops. But why do you thing that there would be induced voltage in these dipoles in the case of the Faraday disk?

I was not considering the Faraday disc, I was considering the Marinov generator and possibly your spinning annulus where the movement is not in a B field.  But you raise an interesting point, if electron acceleration across a brush contact can induce, should this not also apply to the Faraday disc?  The answer must be yes, so the next question should be is there any evidence of the Faraday disc exhibiting anomalous behaviour that could be attributed to this?  Well there are lots of claims for anomalous behaviour that are ridiculed so maybe it is possible.

QuoteWhy do you think that induced voltage would have an effect on the electron spins?

If you replace the spins with current driven loops the current generator then plays its part in the energy audit either as an energy source or sink.

QuoteThe magnet may or may not rotate because there is no interaction of any "counter"-field with those of the magnet's dipoles. This is due to the fact that the magnetic force F=VxB doesn't work, the force being perpendicular to the displacement, so the reaction is not here.
Otherwise rotating a cylindrical magnet in front of another that is coaxial would cause a torque from one to the other. The experiment has been done many times, there is no effect. I'm afraid you're going to go into highly speculative speculation   :).

You have misunderstood what I am trying to convey, it certainly is not a torque on the magnet.

QuoteReplace the permanent magnet of a Faraday disk by a solenoid. I'm sure you will see no induced voltage.

I am not so sure, I think there could be.  Has anyone ever tried it?  It would be a small voltage perhaps too small to register.  It would come from electrons flowing across the outer brush contact where their acceleration creates an unusual electric near-field.  Has anyone ever tried to measure that electric field?
Smudge

https://skemman.is/bitstream/1946/29521/3/KOK_MagneticVectorPotential.pdf

Marinov is on page 51.

--------------
EDIT:
the entire section 5.2.1 Vorticity, Acceleration and Induction of this paper is interesting and poses multiple possible induction methods related tot he vector potential.

Thank you G for that pdf :)

Regards

Mike 8)

My thanks also for that interesting paper.  I can see me studying it closely over the next few days.

Smudge

I find that I already had that, downloaded it 12 months ago but hadn't got round to studying it.  His equation (5.11) is correctly derived in the manner he describes.  He acknowledges that only one component is familiar as the q(vXB) force but fails to point out that his derivation only gives half the value.   So it's not quite right as are other author's papers that just used the identity (5.7) to derive their equations.  I think his Final Words are worth putting here, since it sums up just what this forum is about.
Smudge

FINAL WORDS
To the reader who feels a subtle sense of intuition having been stirred up, I encourage
you to investigate further. Our intuition can be said to be paradoxical, since it
is telling us that "we might know more than we think we know". Even though
the development of science is most often put forth as having been steady linear
step-by-step progression, original ideas often come by conceptual leaps of intuition.
In order to be able to learn, we must admit that we do not know nor understand.
The one who knows and understands everything has nothing to learn. All to often
in the history of science and other fields, have humans carried invalid assumptions
between generations. Until something is discovered or invented, it remains hidden
and might as well be non-existent. Though the path to discovery may be complex
and difficult, the gained knowledge can turn out to be obvious in hindsight, leaving
one to wonder why no one thought of it before. On the other hand, highly original
work is most often with a complete lack of understanding and said to be ahead of
its time, since it can take half a century or more for the community to catch up.
Progress is thus also a matter of timing, with the right ideas being put forth and
appreciated at the right time. With the great accessibility to both education and
scientific literature in modern day and age, along with rapid progress in many areas
of science, it looks as if the time is ripe for a new paradigm. The author's hope has
been that by thinking in terms of the magnetic vector potential, a shift in focus
might come by, leading to a fruitful approach. Mankind is now in desperate need
of new solutions in transportation and energy in order to become independent of
fossil fuels. Even though there are many who have financial interest in the way this
branch of science will develop, research will still, first and foremost, be driven by the
spark of curiosity igniting the wonderful guiding force of inspiration.

Quote from: Smudge on 2019.05.22, 15:17:55
...
FINAL WORDS
..."we might know more than we think we know"
...

Not bad!  :)

I downloaded this paper last year when it came out, put it in a folder to read later, and then forgot about it!  :(
Thanks to Grumpy for putting this valuable link.

Hi Smudge,

I am glad to see that we have made progress. You answered my objection to the question of the different referentials. A is indeed that from the observer's frame of reference, but the gradient of A in space creates a real electric field seen in the frame of reference of the charge (in the same way as E = VxB when using B).
My previous proposal (https://www.overunityresearch.com/index.php?topic=2470.msg74662#msg74662) to prove the concept of mechanical force by virtue of the potential vector should therefore work.

However, the question of the divergence by a factor of 2 between the two methods remains to be resolved. Have you made any progress on this? None on my side!

Quote from: F6FLT on 2019.05.23, 07:13:54
Hi Smudge,

I am glad to see that we have made progress. You answered my objection to the question of the different referentials. A is indeed that from the observer's frame of reference, but the gradient of A in space creates a real electric field seen in the frame of reference of the charge (in the same way as E = VxB when using B).
My previous proposal (https://www.overunityresearch.com/index.php?topic=2470.msg74662#msg74662) to prove the concept of mechanical force by virtue of the potential vector should therefore work.

As should the Marinov generator.  That the Marinov generator can create induction along the velocity vector is possibly a bonus whereas the Faraday disc and your proposal create a sideways force.

QuoteHowever, the question of the divergence by a factor of 2 between the two methods remains to be resolved. Have you made any progress on this? None on my side!

Only the realization that the convective operator by itself is not sufficient to describe motional induction within a B field.  The jury is still out for the non B field.  I am tempted to go back to basics and describe all fields as some sort of interaction of the test particle with the so called virtual particles of space, what I consider to be sub-photons.  Then it might be obvious how the test particle gets its sideways force from that sideways gradient of the A field without actually moving through that gradient.

Smudge
Edit to final sentence.

Quote from: Smudge on 2019.05.23, 14:54:11
...
Only the realization that the convective operator by itself is not sufficient to describe motional induction within a B field.  The jury is still out for the non B field.  I am tempted to go back to basics and describe all fields as some sort of interaction of the test particle with the so called virtual particles of space, what I consider to be sub-photons.  Then it might be obvious how the test particle gets its sideways force from that sideways gradient of the A field without actually moving through that gradient.

Smudge
Edit to final sentence.

Using the potentials or the fields to deal with electromagnetism must give the same results, question of math related to the way these concepts are defined. So I think there is something we missed in the equations, not in the underlying physics.

@F6,

I found this paper by Puthoff buried in my computer.  I think he must have sent it to me way back when we were in contact with each other, as it was in MS Word format.  Interesting that for the long solenoid this alternative approach to EM energy density yields half the magnetostatic energy outside the solenoid where B = 0.

The bad thing is his statement "With regard to applications we first note by Eqns. (9) and (10) that power flux
(and associated momentum transfer) depend on time derivatives ..... and therefore do not attribute momentum transfer to static field distributions."  I take it that his momentum transfer means a force.

Smudge

@Smudge

I understand that since there is no momentum in a static field, you cannot transfert the momentum from one field to another or remove it (so you can't get energy from it). That is what we are observing. But you can have a force without momentum or energy, so for me a force is compatible with a static field. That being said, I don't see very well in practice what it would be.

I was interested in Puthoff a few years ago because of that: https://arxiv.org/pdf/gr-qc/9909037
He tried to give as a physical explanation for the gravity, the change in permeability and permitivity of the vacuum.
This idea was invalidated by more competent people than him. It took me a while to understand why, because at the time I didn't understand relativity well.
In fact, gravity does not change the rhythm of the clocks, but the difference in gravitational potential between two clocks makes them beat at a different rate. It is the gravitational potential, not the field, that is used in the GR equations. This implies an effect that cannot be reduced to physical conditions at the position of the clock. It is surprising that Puthoff was wrong there, whereas this example also shows that the potentials are more fundamental than the fields, which is what Puthoff thinks in the case of electromagnetism.

I agree with what Puthoff says in your paper, in particular with his choice of the Lorentz gauge, by far the most "physical", although I consider him to be a generally very questionable author, not only for this paper I have just quoted, but also because he signed the report on the charlatan Uri Geller affirming his psychic powers of telekinesis, after his evaluation at the Stanford Research Institute in the 1970s. In short, this author may be interesting but he is not reliable.

Quote from: F6FLT on 2019.05.26, 08:53:45
I was interested in Puthoff a few years ago because of that: https://arxiv.org/pdf/gr-qc/9909037
He tried to give as a physical explanation for the gravity, the change in permeability and permitivity of the vacuum.
This idea was invalidated by more competent people than him.

Can you give a link to those please.

QuoteIt took me a while to understand why, because at the time I didn't understand relativity well.
In fact, gravity does not change the rhythm of the clocks, but the difference in gravitational potential between two clocks makes them beat at a different rate.

And Puthoff says just that immediately after equation (11).   As I read it his PV approach is just another way of looking at GR and it gives results that agree with measurements to first order, as do other approaches.

QuoteIt is the gravitational potential, not the field, that is used in the GR equations.  This implies an effect that cannot be reduced to physical conditions at the position of the clock. It is surprising that Puthoff was wrong there

Wrong in what way?  He acknowledges the role of the potentials.  Isn't the gravitational potential at the position of the clock a "physical condition"?

QuoteI agree with what Puthoff says in your paper, in particular with his choice of the Lorentz gauge, by far the most "physical", although I consider him to be a generally very questionable author, not only for this paper I have just quoted, but also because he signed the report on the charlatan Uri Geller affirming his psychic powers of telekinesis, after his evaluation at the Stanford Research Institute in the 1970s. In short, this author may be interesting but he is not reliable.

That may be your view but it is not mine.
Smudge

Quote from: Smudge on 2019.05.26, 14:15:46
Can you give a link to those please.

Puthoff took up an idea of Dicke's from 1957.

"The claims have not been accepted in mainstream physics.
[...]
Mainstream physicists agree that PV is

not viable as a unification of gravitation and electromagnetism
not a "reformulation" of general relativity,
not a viable theory of gravitation, since it violates observational and theoretical requirements."
https://en.wikipedia.org/wiki/Polarizable_vacuum

See wikipedia references for reviews of PV, especially Ibizon.
"The theory predicts a radiation power from a binary system that is 2/3 that predicted by GR, and so incompatible with observed orbital decay rate of PSR 1913+16."
https://arxiv.org/pdf/astro-ph/0302273

Quote
And Puthoff says just that immediately after equation (11).   

Exact, I missed that. But it doesn't change the substance, see below.

Quote
As I read it his PV approach is just another way of looking at GR and it gives results that agree with measurements to first order, as do other approaches.Wrong in what way?  He acknowledges the role of the potentials.  Isn't the gravitational potential at the position of the clock a "physical condition"?That may be your view but it is not mine.
Smudge

The twin paradox is solved not only by special relativity, but also by general relativity. In this case the twin who turns around is subjected to centripetal acceleration, and by the principle of acceleration/gravity equivalence, its clock is seen by the twin at rest in the same way as if it were subjected to the same gravity value as the acceleration. Thus the difference in "gravitational" potential between the two will determine the clocks time lag.
How does Puthoff explain this case, and the compatibility of PV with the principle of equivalence in general?

Quote from: F6FLT on 2019.05.26, 17:42:17
See wikipedia references for reviews of PV, especially Ibizon.
"The theory predicts a radiation power from a binary system that is 2/3 that predicted by GR, and so incompatible with observed orbital decay rate of PSR 1913+16."
https://arxiv.org/pdf/astro-ph/0302273

So Puthoff employs Ibison to create a critique of the PV cosmology and that makes Puthoff "not reliable" (Your words).
Smudge

Puthoff's Earthtech organisation should be congratulated for their diligent investigations into excess energy claims
https://earthtech.org/excess-energy-claims/ (https://earthtech.org/excess-energy-claims/)
And this man Puthoff is "not reliable"?
Smudge

Quote from: Smudge on 2019.05.27, 16:45:07
So Puthoff employs Ibison to create a critique of the PV cosmology and that makes Puthoff "not reliable" (Your words).
Smudge

It is only one of several elements. You missed the fact that I made it clear, that he was completely fooled by a magician and asserted a psychic power of telekinesis in Uri Geller, who turned out to be a charlatan. It is a symptom of those who absolutely want to see in reality, what they dream.

An alternative theory is only needed if it models observations that were not, if it can predict new things, or if it synthesizes several theories into a simpler and more comprehensive theory.

What new things does Puthoff predict that we could experimentally verify or that would not be yet explained by conventional theories?

Quote from: Smudge on 2019.05.27, 16:56:47
Puthoff's Earthtech organisation should be congratulated for their diligent investigations into excess energy claims
https://earthtech.org/excess-energy-claims/ (https://earthtech.org/excess-energy-claims/)
And this man Puthoff is "not reliable"?
Smudge

Again: they say their mission is "Shaping the future by innovating breakthroughs", where is their "breakthrough" technology? Have they at least an experimental proof of a new concept?

Quote from: Smudge on 2018.06.03, 16:00:00
I like Tinman's idea for using a plastic food bowl for a quick budget experiment.  His idea of using two horseshoe magnets should work OK but of course the output is low voltage DC.  I think an AC experiment is worthwhile because if this really works and is OU then it will need AC generation.  This is easily achieved by using a soft ferromagnetic ring core driven with AC.  Many of you will have small ring cores that would fit into a plastic drinking glass, so that could be a route to a budget experiment, see images below.  And you will have oscilloscopes that can easily measure low voltage AC to compare with the input.  If it works there should be low voltage AC output that is independent of frequency, in phase with the input current and proportional to rotation speed.  If that is achieved it shows there to be an unusual form of induction that could be OU in that output energy is not derived from the input to the coil, but comes from whatever space energy keeps electrons perpetually spinning and orbiting.  Anyone up for this?
Smudge

Well i took the time tonight to build and run the experiment,which i always intended to do when i got some free time. But upon coming back here and re-reading the experiment parameters,it would seem i screwed it up a bit.

I used just an open rod magnet,where north was facing one brush,and south was facing the other brush.
So in my experiment,the magnetic field was not looped,it was open.

I will post the video here,as i have already uploaded it.
Although it may work like a homopolar generator,it is still quite surprising to see a voltage across the shorted ring.
With a homopolar generator,we have one brush on the inner axis,and one on the outer circumference.
But with this setup,we have both brushes around the circumference.

I will replace the PM with a toroid tomorrow,run AC through it,and see if we get a voltage that way.
As the voltage is very low now with the fields cutting the conductor,i have my doubts that a voltage will be produced when those fields are contained within the toroid--but we will see tomorrow.

https://www.youtube.com/watch?v=7fu2cYlTcvg


Brad

Thank you Brad for your interest in this.  With normal "flux cutting" the induction is at right angles to the movement, but here the induction is along the movement.  That in itself is different from what is normally accepted motional induction.  I look forward to your results with a toroidal core.  You will probably need to drive the toroid to its saturation limit to get maximum core flux.  If I am right, with AC in the toroid and no movement you should get zero voltage.  But with the slip-ring rotating you should get a small AC induction which is proportional to the RPM.  Of course that small voltage can be stepped up using a transformer so that it gets to a useful level.  And that might allow some power measurements looking for signs of OU.  Interesting times ahead possibly.
Smudge

Hi Brad,

A slight magnetic field gradient possibly exists along the cylinder between the two sliding contacts and explains the voltage. I think the reason for your voltage should be the same as mine here where this effect is enhanced by the asymmetry of the field along the cylinder: https://www.overunityresearch.com/index.php?topic=3738.0

By moving the position of one of your contacts a little along the cylinder, towards or away from the motor, the voltage should drop to zero. Could you check this hypothesis?

In any case, the magnetic flux is always looped through the air, between the poles of the magnet.

Quote from: F6FLT on 2019.06.04, 06:28:48
Hi Brad,

A slight magnetic field gradient possibly exists along the cylinder between the two sliding contacts and explains the voltage. I think the reason for your voltage should be the same as mine here where this effect is enhanced by the asymmetry of the field along the cylinder: https://www.overunityresearch.com/index.php?topic=3738.0

By moving the position of one of your contacts a little along the cylinder, towards or away from the motor, the voltage should drop to zero. Could you check this hypothesis?

In any case, the magnetic flux is always looped through the air, between the poles of the magnet.

Ok,so i have this right--
Leave one brush as is,and move the other either toward the front of the drum,or to the rear of the drum,which would mean back toward the motor in the later.
So we are sliding the brush across the width of the drum--correct?.

I will give it a shot tomorrow  O0


Brad

Quote from: TinMan on 2019.06.04, 14:08:37
...
Leave one brush as is,and move the other either toward the front of the drum,or to the rear of the drum,which would mean back toward the motor in the later.
So we are sliding the brush across the width of the drum--correct?.
...

That's right, that's the idea.
In my opinion, it is the distance along the drum between the two circumferences determined by the sliding contacts which are not perfectly aligned, linked to a magnetic field gradient, that makes the voltage.
So if you could also make the sliding contacts sharper, more punctual, it would be better, otherwise the wide surface makes it unclear by which point (or points) the current really escapes from the width of the drum to the contacts.

Quote from: F6FLT on 2019.06.05, 07:31:18
That's right, that's the idea.
In my opinion, it is the distance along the drum between the two circumferences determined by the sliding contacts which are not perfectly aligned, linked to a magnetic field gradient, that makes the voltage.
So if you could also make the sliding contacts sharper, more punctual, it would be better, otherwise the wide surface makes it unclear by which point (or points) the current really escapes from the width of the drum to the contacts.

I ran the test.
I put the edge of 1 brush on the inner edge of the drum,and the edge of the other brush on the outer edge of the drum.
This resulted in a much higher voltage being produced across the brushes,some of which could be attributed to a slightly higher RPM due to less brush friction.

I will post the video later on tonight.


Brad

There are classic drum HPGs, I mean where the current is caught between the center of the axis and the periphery of the drum, capable of producing hundreds of amperes. In these devices, the external contact is made by a multitude of electrodes connected together, in contact with the drum all around.
If current were to flow significantly between these sliding peripheral contacts, which your setup suggests, I think it would have been seen, it would cause significant losses. That's why I don't believe too much in a current generated along the circumference. Beware of possible artifacts, you have to be very careful.

Quote from: F6FLT on 2019.06.05, 08:41:29
There are classic drum HPGs, I mean where the current is caught between the center of the axis and the periphery of the drum, capable of producing hundreds of amperes. In these devices, the external contact is made by a multitude of electrodes connected together, in contact with the drum all around.
If current were to flow significantly between these sliding peripheral contacts, which your setup suggests, I think it would have been seen, it would cause significant losses. That's why I don't believe too much in a current generated along the circumference. Beware of possible artifacts, you have to be very careful.

If we had the scope just across the brushes,and there was no cap,then i would agree that artifacts could be a possibility. But as we have a 35uF cap across the brushes,and the scope is across the cap,i would think that a voltage built up across that cap would mean a current flow to that cap. And once the motor is stopped,the cap is bled down,and then recharged with the opposite polarity when the motor is started in the opposite direction,or by flipping the magnet poles around on the fly.

There are losses here at play though,and they are of course eddy current losses,as the copper ring gets hot after a short time of spinning.

The current is also very low (with my setup at least),where a 100 ohm resistor across the cap will reduce the voltage by half. But even so,i am surprised there is any at all.
It is without doubt very inefficient,as most of the motors energy is converted to heat through eddy currents in the copper ring.

https://www.youtube.com/watch?v=OCA01hzLYOE


Brad

I don't buy into F6's electrons dragged through a magnetic gradient.  They would have to be spin polarized and in Cu they are not.  Collisions with the atoms ensures that can't happen.  If the slip-ring were ferromagnetic then that would be a different ball game.

I am still hooked on movement through the magnetic vector potential A field as a possibility.  This A field forms concentric rings around the magnet axis.  I would like to offer a slightly different experiment if Brad is willing to continue.  Remove the magnet from inside the slip-ring and hold magnets near.  One that interests me is shown in the attached image.  For this you need a long bar magnet, perhaps put two or three smaller ones together end to end.  The magnet center needs to be as close as possible to the brush contacts.  The eddy current problem should be reduced because with long magnets there is virtually zero B field close to the middle section.  If you still get induced voltage then I think that is fairly strong evidence for this A field effect.   And I think it will not work as a motor so that non-reciprocity bodes well for this offering the potential for OU operation.  If Brad does get induction I think he can claim a world's first for a definitive experiment of this kind.
Smudge 

Quote from: Smudge on 2019.06.05, 10:49:53
I don't buy into F6's electrons dragged through a magnetic gradient.  They would have to be spin polarized and in Cu they are not.  Collisions with the atoms ensures that can't happen.  If the slip-ring were ferromagnetic then that would be a different ball game.

I am still hooked on movement through the magnetic vector potential A field as a possibility.  This A field forms concentric rings around the magnet axis.  I would like to offer a slightly different experiment if Brad is willing to continue.  Remove the magnet from inside the slip-ring and hold magnets near.  One that interests me is shown in the attached image.  For this you need a long bar magnet, perhaps put two or three smaller ones together end to end.  The magnet center needs to be as close as possible to the brush contacts.  The eddy current problem should be reduced because with long magnets there is virtually zero B field close to the middle section.  If you still get induced voltage then I think that is fairly strong evidence for this A field effect.   And I think it will not work as a motor so that non-reciprocity bodes well for this offering the potential for OU operation.  If Brad does get induction I think he can claim a world's first for a definitive experiment of this kind.
Smudge

Just so happens that i have brand new long neo bar magnets that will do the trick.

I'll put on my jumper,and head on out to the workshop now,and give it a try

Back soon.

Quote from: TinMan on 2019.06.05, 11:24:08
Just so happens that i have brand new long neo bar magnets that will do the trick.

I'll put on my jumper,and head on out to the workshop now,and give it a try

Back soon.

Bright idea! Especially if you pick the current right in the middle.

If you have a long bar magnet, you could also try this:
https://www.overunityresearch.com/index.php?topic=2470.msg74662#msg74662 (if there is a voltage, this would be the first experimental evidence of a HPG with the rotating conductor outside the magnetic field).

Quote from: F6FLT on 2019.06.05, 15:20:06
Bright idea! Especially if you pick the current right in the middle.

If you have a long bar magnet, you could also try this:
https://www.overunityresearch.com/index.php?topic=2470.msg74662#msg74662 (if there is a voltage, this would be the first experimental evidence of a HPG with the rotating conductor outside the magnetic field).

Not sure i understand there F6,as the B field exists all along the length of a magnet or solenoid,and so a voltage would indeed be produced across the disc. The magnetic field would cut the disc just as that of a HPG.


Brad

Quote from: TinMan on 2019.06.05, 16:20:00
Not sure i understand there F6,as the B field exists all along the length of a magnet or solenoid,and so a voltage would indeed be produced across the disc. The magnetic field would cut the disc just as that of a HPG.


Brad

There is no B field outside a long solenoid (in fact there is a weak residual field if the solenoid is not infinite). See here explanation, § "Outside" (https://web.ua.es/docivis/magnet/solenoid.html). A long cylinder magnet is exactly like a long solenoid carrying a DC current.
In the proposed experiment, the disk has a hole, and B=0 at the position of the remaining part. The magnetic flux is only crossing the hole, there is no flux through the metal part.

The Lorentz force applying to the disk electrons is the explanation of the Faraday disk. Here it can't apply. Nevertheless I expect a current. A HPG without magnetic flux through the rotating conductor would be a great innovation. I haven't seen it anywhere else except in the Marinov's setup which is a proof of concept more complicated than mine.

Quote from: F6FLT on 2019.06.06, 07:18:33
There is no B field outside a long solenoid (in fact there is a weak residual field if the solenoid is not infinite). See here explanation, § "Outside" (https://web.ua.es/docivis/magnet/solenoid.html). A long cylinder magnet is exactly like a long solenoid carrying a DC current.
In the proposed experiment, the disk has a hole, and B=0 at the position of the remaining part. The magnetic flux is only crossing the hole, there is no flux through the metal part.

The Lorentz force applying to the disk electrons is the explanation of the Faraday disk. Here it can't apply. Nevertheless I expect a current. A HPG without magnetic flux through the rotating conductor would be a great innovation. I haven't seen it anywhere else except in the Marinov's setup which is a proof of concept more complicated than mine.

Ah,ok
Looking at your sketch again,i see the magnet has to be a cylinder and not a solid rod magnet.

I assume this is the difference ?.


Brad

Quote from: TinMan on 2019.06.06, 09:38:14
Ah,ok
Looking at your sketch again,i see the magnet has to be a cylinder and not a solid rod magnet.

I assume this is the difference ?.


Brad

The magnet is a solid rod magnet. By "cylinder" I meant a cylinder that was not hollow, that was not just a drum.
But maybe I didn't understand your question.

Actually, it doesn't matter if there's a hole in the magnet. For example, the rod could be composed of many disc magnets or ring magnets. The important point is that the magnetization is along the axis.

Quote from: F6FLT on 2019.06.06, 11:13:44
The magnet is a solid rod magnet. By "cylinder" I meant a cylinder that was not hollow, that was not just a drum.
But maybe I didn't understand your question.

Actually, it doesn't matter if there's a hole in the magnet. For example, the rod could be composed of many disc magnets or ring magnets. The important point is that the magnetization is along the axis.

Ok,well then i still do not understand as to why this setup of yours will not work just like a HPG,as the magnetic flux will still be cutting through the disk.
I can tell you right now that it will work as a HPM,so therefor it should also work as a generator.

If we were to use a long solenoid,and where the copper disk was a thin secondary coil,then passing an AC current through the solenoid will result in the thin secondary being induced.


Brad

Quote from: TinMan on 2019.06.06, 12:24:16
Ok,well then i still do not understand as to why this setup of yours will not work just like a HPG,as the magnetic flux will still be cutting through the disk.
...

However, I gave the key to this understanding: Lorentz's force. Electrons are only subjected to this force if they bathe in the magnetic field. No B field through the disc between the two sliding contacts, so no radial Lorentz's force on the electrons.
And yet there should be one.

QuoteIf we were to use a long solenoid,and where the copper disk was a thin secondary coil,then passing an AC current through the solenoid will result in the thin secondary being induced.

In this case the force on the electrons is tangential, not radial, and due to dA/dt, not to the Lorentz force.
Nevertheless it's the equivalent for the mechanical case that my setup should show: force without a B field at the position of the electrons.

Quote from: F6FLT on 2019.06.06, 13:06:43
However, I gave the key to this understanding: Lorentz's force. Electrons are only subjected to this force if they bathe in the magnetic field. No B field through the disc between the two sliding contacts, so no radial Lorentz's force on the electrons.

Well you've lost me,as i see a clear B field through the disk,and the Lorents force will be at right angles to the B field.
It seems no different to that of a HPG to me.
See pic below.

Quote from: TinMan on 2019.06.06, 16:58:03
Well you've lost me,as i see a clear B field through the disk,and the Lorents force will be at right angles to the B field.
It seems no different to that of a HPG to me.
See pic below.

You have shown a poor representation, probably from weak magnets (AlNiCo or ferrite which have more leaks), and the rod is not long enough compared to the size of the disc.
See what you will have with a long neodymium rod:

(https://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=31973;image)

See also : https://www.quora.com/Why-is-there-no-magnetic-field-outside-the-current-carrying-solenoid

Quote from: F6FLT on 2019.06.07, 08:00:55
You have shown a poor representation, probably from weak magnets (AlNiCo or ferrite which have more leaks), and the rod is not long enough compared to the size of the disc.
See what you will have with a long neodymium rod:

(https://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=31973;image)

See also : https://www.quora.com/Why-is-there-no-magnetic-field-outside-the-current-carrying-solenoid

Ah,ok.
Well we would need a really long magnet for that,1 which i do not have.


Brad

For those here with an understanding of the flux field around a solenoid or long cylindrical PM, where will the energized wire go shown in the illustration below and why?

Regards,
Pm

Quote from: partzman on 2019.06.08, 12:36:48
For those here with an understanding of the flux field around a solenoid or long cylindrical PM, where will the energized wire go shown in the illustration below and why?

Regards,
Pm

What do you mean by !energized wire!  ?

I think he means a wire carrying a current into the page as denoted by the cross or plus sign in the circle.  The wire will follow Fleming's LH rule.
Smudge

Quote from: Smudge on 2019.06.08, 14:08:54
I think he means a wire carrying a current into the page as denoted by the cross or plus sign in the circle.  The wire will follow Fleming's LH rule.
Smudge

Well in that case we need to know--
1-what is it that is trying to be achieved ?
2-where is the position of the second current carrying wire ?

Quote from: TinMan on 2019.06.08, 11:19:13
Ah,ok.
Well we would need a really long magnet for that,1 which i do not have.


Brad

Your magnet is long enough. Simply use a smaller disc, for example with a rim around the magnet at a distance approximately equal to the radius of the rod.
The proof of concept is obtained if you measured about the same voltage between the center of the magnet and its periphery (functioning with the magnetic field through the disk, classical "Faraday's disk generator"), as between the periphery of the magnet at the position of the disk and the rim of the disk (functioning with the vector potential and without magnetic field through the disk).

Quote from: TinMan on 2019.06.09, 00:52:13
Well in that case we need to know--
1-what is it that is trying to be achieved ?
2-where is the position of the second current carrying wire ?

I find it interesting that no one to date has ventured a guess!  Or done the experiment!  With all due respect, I see much theory being discussed here so I would think this should be easy to analyze or answer.

My point in posting this was to have the results explained.  I have my own idea but maybe it's so simple that I just don't "get it".

There is only one current carrying wire positioned in close proximity of the PM with the ability to move somewhat freely.  The ends of this wire simply connect to a DC power supply.

Regards,
Pm

I did the experiment and like mentioned by Smudge, the wire will follow Fleming's LH rule, so the field will
be like drawn and the both fields will repel causing the wire to move to the right.

Itsu

Quote from: Itsu on 2019.06.09, 15:21:08
I did the experiment and like mentioned by Smudge, the wire will follow Fleming's LH rule, so the field will
be like drawn and the both fields will repel causing the wire to move to the right.

Itsu

Will the wire also move up or down?

Am I out out of the park here to suggest the magnet will move in relationship to the wire, like in John Bedini's zero force motor?

https://www.youtube.com/watch?v=w-GZerEwObo (https://www.youtube.com/watch?v=w-GZerEwObo)

Ron

Quote from: Itsu on 2019.06.09, 15:21:08
I did the experiment and like mentioned by Smudge, the wire will follow Fleming's LH rule, so the field will
be like drawn and the both fields will repel causing the wire to move to the right.

Itsu

ooops,   my bad,   i did not check the polarity of the magnet,  turns out in the above post it was South up instead of North.
Doing it again making sure North is up,  the wire is attracted to the magnet, so to the left......

Laying the magnets flat (horizontal) on the bench does not show any movement left or right,   meaning up or down when put vertical.

Sorry for the mistake,  Itsu

Quote from: Itsu on 2019.06.09, 15:21:08
I did the experiment and like mentioned by Smudge, the wire will follow Fleming's LH rule, so the field will
be like drawn and the both fields will repel causing the wire to move to the right.

Itsu

Itsu,

Was your wire movement restricted?  Did the wire only move to the right?

Pm

Quote from: Itsu on 2019.06.09, 20:18:40

ooops,   my bad,   i did not check the polarity of the magnet,  turns out in the above post it was South up instead of North.
Doing it again making sure North is up,  the wire is attracted to the magnet, so to the left......

Laying the magnets flat (horizontal) on the bench does not show any movement left or right,   meaning up or down when put vertical.

Sorry for the mistake,  Itsu

Itsu,

I missed this post of yours while I was creating my previous post so my question now is, where did the wire end up in relation to the magnet or where did it attach to the magnet?

Pm

PM,

the wire is a cliplead wire about 30cm long fixed on both sides hanging in a loop.
Magnet is a stack of ceramic disk magnets 5cm long.

at about 0.5cm from the magnet stack, the wire will move about 3mm, almost touching the magnet.

Redoing the horizontal test shows a very slight movement to the left (north at left).

Using about 7V from my PS going into current limitation.

Itsu

Quote from: Itsu on 2019.06.09, 20:40:02
PM,

the wire is a cliplead wire about 30cm long fixed on both sides hanging in a loop.
Magnet is a stack of ceramic disk magnets 5cm long.

at about 0.5cm from the magnet stack, the wire will move about 3mm, almost touching the magnet.

Redoing the horizontal test shows a very slight movement to the left (north at left).

Using about 7V from my PS going into current limitation.

Itsu

Itsu,

Thanks for doing the experiment.  I have included a link to a video of my results done a few years back and I'm curious to see if you get the same results with your PM assembly. 

https://www.youtube.com/watch?v=F2JFDpTE_ls

My question is, why the apparent attraction to the bloch wall?

Pm

Quote from: ronee on 2019.06.09, 19:05:03
Will the wire also move up or down?

Am I out out of the park here to suggest the magnet will move in relationship to the wire, like in John ######'s zero force motor?

https://www.youtube.com/watch?v=w-GZerEwObo (https://www.youtube.com/watch?v=w-GZerEwObo)

Ron

There is nothing special about the zero force motor.
It's simply the magnet being repelled from the like field of the solenoid,and attracted to the unlike field of the solenoid.


Brad

Quote from: partzman on 2019.06.09, 22:00:26
Itsu,

Thanks for doing the experiment.  I have included a link to a video of my results done a few years back and I'm curious to see if you get the same results with your PM assembly. 

https://www.youtube.com/watch?v=F2JFDpTE_ls

My question is, why the apparent attraction to the bloch wall?

Pm

First off,there is no !bloch wall! as you call it.

As far as your current carrying conductor go's--
Im not sure as to why it would choose a set side of the magnet,depending on which way the current flows through the conductor.

I will have to experiment with that tonight.
I would think it has something to do with the field spin of the PM,and the field spin around the conductor.
A PMs field spin is concentrated at the center of the two poles,so this may answer that.


Brad

Quote from: partzman on 2019.06.09, 22:00:26
Itsu,

Thanks for doing the experiment.  I have included a link to a video of my results done a few years back and I'm curious to see if you get the same results with your PM assembly. 

https://www.youtube.com/watch?v=F2JFDpTE_ls

My question is, why the apparent attraction to the bloch wall?

Pm

Ok,this is indeed odd,and maybe some one here can explain it,but i cant.

QuoteMy question is, why the apparent attraction to the bloch wall?

I have found it depends on which way your PM is facing as to whether the current carrying conductor is attracted to the center of the magnet,or repelled away from it.

So far i have found the following

1-the current carrying conductor is repelled by both the north and south fields of the PM,even though the current flow direction remains the same.

2-if the current carrying conductor is attracted to the center of the magnet,you can either invert the current flow through the inductor,or spin the magnet 180* to make the conductor become repelled by the center of the magnet.

3-The conductor is always repelled by the center on one side of the PM,but attracted to the center on the other side of the PM.

I think this effect is caused by the field within the center of the PM,and not the outer field of the PM.

If we look at the picture below,where we see the field of the PM,and the field of a current carrying conductor,and we take note of the PMs field inside the magnet,and disregard the field around the PM,we see that if the conductor is on one side of the PM,the magnetic field flow is the same direction. If we place the conductor on the other side of the PM,we see the fields flow is in the opposite direction. So on one side of the PM the conductor would be attracted to the magnet,and on the other side it would be repelled. And as i have stated before,the magnetic field is at it's most concentrated at the center of the magnet,not at each pole.

This would explain what we see in my video below.

https://www.youtube.com/watch?v=m5SKYiSxSBk

Im calling this the Partzman effect  O0


Brad

Quote from: TinMan on 2019.06.10, 12:10:47
...If we place the conductor on the other side of the PM,we see the fields flow is in the opposite direction. So on one side of the PM the conductor would be attracted to the magnet,and on the other side it would be repelled. And as i have stated before,the magnetic field is at it's most concentrated at the center of the magnet,not at each pole.
...

The wire moves through the field gradient and "falls" to the position of minimum magnetic potential energy. No surprise.

Quote from: F6FLT on 2019.06.10, 12:26:58
The wire moves through the field gradient and "falls" to the position of minimum magnetic potential energy. No surprise.

F6,

Would you show more clearly by graphic means what you are saying?  It appears the wire does more than "fall" as it seems to be "pinned" to it's position!

Regards,
Pm

Quote from: TinMan on 2019.06.10, 12:10:47
Ok,this is indeed odd,and maybe some one here can explain it,but i cant.

I have found it depends on which way your PM is facing as to whether the current carrying conductor is attracted to the center of the magnet,or repelled away from it.

So far i have found the following

1-the current carrying conductor is repelled by both the north and south fields of the PM,even though the current flow direction remains the same.

2-if the current carrying conductor is attracted to the center of the magnet,you can either invert the current flow through the inductor,or spin the magnet 180* to make the conductor become repelled by the center of the magnet.

3-The conductor is always repelled by the center on one side of the PM,but attracted to the center on the other side of the PM.

I think this effect is caused by the field within the center of the PM,and not the outer field of the PM.

If we look at the picture below,where we see the field of the PM,and the field of a current carrying conductor,and we take note of the PMs field inside the magnet,and disregard the field around the PM,we see that if the conductor is on one side of the PM,the magnetic field flow is the same direction. If we place the conductor on the other side of the PM,we see the fields flow is in the opposite direction. So on one side of the PM the conductor would be attracted to the magnet,and on the other side it would be repelled. And as i have stated before,the magnetic field is at it's most concentrated at the center of the magnet,not at each pole.

This would explain what we see in my video below.

https://www.youtube.com/watch?v=m5SKYiSxSBk

Im calling this the Partzman effect  O0


Brad

Brad,

Thanks for doing your experiment and creating the video! 

Your video confirms IMO that perhaps we do not have the magnetic field properly defined around the cylindrical PM or solenoid coil.

Regards,

Pm

Below is a photo that best explains my answer to what is happening with this simple PM/wire experiment.  This was taken from this website-

http://www.ferrocell.us/images/2-anothermoebius3.jpg

and here is the caption for the pix-

"Cylinder magnet on top of 50mm rear lit cell using a
small incandescent lamp. Note indent right, center
and bulge left center."

Most of our current depictions of flux lines in a bar or cylindrical PMs magnetized on their long dimension show continuous lines from N to S.  This is all wrong IMO as these simple experiments seem to indicate as is also seen in the backlit ferrocell views.  IOW, the N flux lines return to the center of the PM as do the S flux lines.  Also IMO, these lines are anything but straight as is always depicted.

If this is the case, then the mystery is solved as to why the wire attracts to the positions it does by considering the attraction and repulsion of the respective flux fields.

Regards,
Pm     

Quote from: F6FLT on 2019.06.10, 12:26:58
The wire moves through the field gradient and "falls" to the position of minimum magnetic potential energy. No surprise.

Not sure if you watched my video i posted,but your answer is incorrect.
The wire can either be attracted to,or repelled from the center of the magnet.
There is no !falling! .


Brad

Quote from: partzman on 2019.06.10, 13:44:57
Below is a photo that best explains my answer to what is happening with this simple PM/wire experiment.  This was taken from this website-

http://www.ferrocell.us/images/2-anothermoebius3.jpg

and here is the caption for the pix-

"Cylinder magnet on top of 50mm rear lit cell using a
small incandescent lamp. Note indent right, center
and bulge left center."

Most of our current depictions of flux lines in a bar or cylindrical PMs magnetized on their long dimension show continuous lines from N to S.  This is all wrong IMO as these simple experiments seem to indicate as is also seen in the backlit ferrocell views.  IOW, the N flux lines return to the center of the PM as do the S flux lines.  Also IMO, these lines are anything but straight as is always depicted.

If this is the case, then the mystery is solved as to why the wire attracts to the positions it does by considering the attraction and repulsion of the respective flux fields.

Regards,
Pm     

Well whats the chances of that.
I see a figure 8 magnetic field  ;)

Way back in 2015 i was grilled by the ex-spurts when i stated that the magnetic field around a PM is in the form of a figure 8. This is when i spent many months mapping field lines around PM.

Quote from thread at OU.com -->It seems all well and good for the(so called) !know all! scientist to place lines around a magnet to represent the magnetic field,but when i draw my lines that separate the TWO different fields around a magnet ,and show where each individual field is strongest,every one of the guru's say thats crap. Well,to bad,my mapping of the two different fields dose form a figure 8 pattern relative to the magnetic field strength of each individual field-north and south.

The whole thread here-->

https://overunity.com/14974/magnet-myths-and-misconceptions/870/

Now,after all these years of being told i was wrong,look what turns up  C.C
Aint that a hoot  ;D

Partzman
I agree with you 100%
In fact,i agreed with you long before you posted this lol.

This is a moment to remember for sure.
Everything changes now  O0


Brad

Brad,

Yes, if what we see and speculate regarding the field lines is correct, it begs the question of can be benefit in any way from this knowledge in the various PM motor and generator designs?

BTW, Ken Wheeler has been preaching along these lines for years for those who may not know.

Regards,
Pm

Why does this "8" not show up with the iron fillings test?

Itsu

Quote from: Itsu on 2019.06.10, 19:22:34
Why does this "8" not show up with the iron fillings test?

Itsu

Itsu,

I'm not qualified to answer that however, look closely at the white space that is between the iron filings attached to the PM and the distant iron filings!  What do you see?  The iron filings also show flux lines that are discontinuous which we are told do not exist under normal conditions like this.  Also, how does one justify the wire movement and position based on the iron filing flux indication?

Regards,
Pm

Quote from: TinMan on 2019.06.10, 13:48:11
Not sure if you watched my video i posted,but your answer is incorrect.
The wire can either be attracted to,or repelled from the center of the magnet.
There is no !falling! .


Brad

Well if we visualize the magnetic field of the magnet(outside the magnet from pole to pole) and then the field of the energized wire(the circumference of the wire field in the space between the wire and the magnet), I can see the magnet field in opposition to the wire field in one polarity of input and attraction in opposite polarity input.  The sides of the cylinder magnet do emanate pole fields outward from each half on either side of length center. 

The floating and attraction effect should be easy to visualize as the poles in the circular field around the wire are either in attraction or repulsion to the poles of the magnet, like just bringing in another magnet parallel to the first magnet where its attracted, then flipping the magnet for repulsion

So lets say you have the wire on this side of center and polarity to do the flip around effect. Flux is in a general outward direction(some at angles toward the other pole closer to the middle of the magnet),so when we apply the current when the wire is off center, the wire is going to move in the same direction as if it were started off in front of the pole, around the say N field that is emanating out of the N end of the magnet.  Now we move the wire to the other side of center and now it reacts to S pole field that emanates around that half of the magnet.

If you make a window motor with just 1 wire as you show, just on 1 side of the rotor around the magnet, it will be a similar effect I believe. But still a good effect to see the wire having the ability to do what it is doing without the restrictions of just going around in a predetermined circle.



Mags

Quote from: partzman on 2019.06.10, 13:00:00
F6,

Would you show more clearly by graphic means what you are saying?  It appears the wire does more than "fall" as it seems to be "pinned" to it's position!

Regards,
Pm

Hi PM,

It is "pinned" when it is on the right side (the one where its potential energy is minimal), otherwise it is sent back to the other side.
You can see it in Brad's video: https://www.youtube.com/watch?v=m5SKYiSxSBk&feature=youtu.be&t=121

Quote from: TinMan on 2019.06.10, 13:48:11
...
The wire can either be attracted to,or repelled from the center of the magnet.
There is no !falling! .
...

Hi Brad,

"falling" is just used by analogy with gravitational potential. The same reason that will make a balloon roll from the top to the bottom of the hill, makes the wire move from one side of the magnet to the other: the moving object loses the potential energy acquired by its starting position.

Quote from: partzman on 2019.06.10, 13:44:57
...
http://www.ferrocell.us/images/2-anothermoebius3.jpg
...     

If it's not a simple mechanical effect on the surface, I don't believe for a second that this magnet would be a simple cylindrical magnet axially magnetized. It would be more like the attached file.
And if it's a mechanical effect, either due to the magnet weight or to magnetic forces on the magnet because of ferromagnetic objects in the surrounding, no conclusion can be drawn about the magnetism.

Quote from: partzman on 2019.06.10, 20:53:12
Itsu,

I'm not qualified to answer that however, look closely at the white space that is between the iron filings attached to the PM and the distant iron filings!  What do you see?  The iron filings also show flux lines that are discontinuous which we are told do not exist under normal conditions like this.  Also, how does one justify the wire movement and position based on the iron filing flux indication?

Regards,
Pm

Do not confuse the map with the territory. Iron filings are not the field lines, they line up along the field lines. But there are grain irregularities and also friction forces on the surface that make there are never only continuous lines while the field lines are.
The field lines are not even a physical reality, at least at the scale of iron filings. They are just mathematical lines of magnetic equipotentials.
If the field lines were not continuous, the magnetic force would not be conservative, and SMOTs would have been successful for years.

Quote from: F6FLT on 2019.06.11, 08:18:00
Hi PM,

It is "pinned" when it is on the right side (the one where its potential energy is minimal), otherwise it is sent back to the other side.
You can see it in Brad's video: https://www.youtube.com/watch?v=m5SKYiSxSBk&feature=youtu.be&t=121


Hi Brad,

"falling" is just used by analogy with gravitational potential. The same reason that will make a balloon roll from the top to the bottom of the hill, makes the wire move from one side of the magnet to the other: the moving object loses the potential energy acquired by its starting position.

So here is what we have.
Pic 1--If we see the current flowing into the page through the conductor,the conductor is pushed up away from the magnet. It is not attracted to either the north or south field,but repelled by both.
Following the green lines,the conductor can travel either way around the magnet,and then be pulled up toward the center of the magnet.

Pic two--
By flipping the magnet over,we see the conductor pulled straight down to the magnets center.
We also get the same effect by swapping the polarity of the current flow,and leaving the magnet as it is.

So how is it that the conductor is repelled by both the north field and the south field ?.

The last 2 pictures make more sense.


Brad

Quote from: Itsu on 2019.06.10, 19:22:34
Why does this "8" not show up with the iron fillings test?

Itsu

Iron filings are no good at showing field lines as such,as each filing particle becomes a temporary magnet it self,and so just stick together how ever they like.


Brad

Brad, Itsu, PM, F6, Guys,
You are comparing apples to oranges.  Field lines are lines of force, and here we have two different lines of force.  First off the B field lines show the direction of a force that would apply to an isolated magnetic N pole.  Such an isolated N pole would be repelled from the N pole of the magnet and would travel along one of those field lines to reach the S pole of the magnet.  We don't have isolated poles but you can do an experiment that will show this effect.  Make a long thin magnet by magnetizing a length of rigid piano wire, then suspend this from a ceiling so that it hangs down as a pendulum with the S pole at the suspension point, a pendulum that can follow the surface of a cone.  Then place a small magnet directly below it to see the N pole of the pendulum repelled from the N pole of the magnet.  You will see that N pole follow a curved trajectory as depicted in the rather poor image attached.   That is showing you the B field line.  It is a line of force and if you integrate that line from The N pole of the magnet to the S pole you get the energy (force times distance) or work done.  So from that starting position there is that potential energy available.  It is not just some vague potential, it is potential energy.  If you started somewhere else on the force line you would get less potential energy.  If you wanted to you could draw other "field" lines like lines of equipotential energy.  Iron filings are magnetized by the B field to become tiny magnets and they do link up to form strings that follow those B lines.

Brad's figure of eight field lines are a different field.  They are not the force on an isolated N pole, they are the force on a line element of current.  Brad is correct to see them as field lines, and to call them magnetic field lines as clearly they come from magnetism.  They need a different terminology.  We already have magnetic A fields, magnetic B fields and magnetic H fields.  Perhaps we should call them F8 fields because of that figure of 8.  Or maybe we could use FLH fields as they obey Fleming's LH rule.  Or why not just magnetic F field?  Since by Fleming's LH rule the force is always at right angles to the B field, you can easily plot the F field lines as shown in the second image.   And being lines of force there is potential energy available so you could also plot lines of equipotential energy of you wanted to.
Smudge

Quote from: TinMan on 2019.06.11, 10:45:24
So here is what we have.
Pic 1--If we see the current flowing into the page through the conductor,the conductor is pushed up away from the magnet. It is not attracted to either the north or south field,but repelled by both.
Following the green lines,the conductor can travel either way around the magnet,and then be pulled up toward the center of the magnet.

Pic two--
By flipping the magnet over,we see the conductor pulled straight down to the magnets center.
We also get the same effect by swapping the polarity of the current flow,and leaving the magnet as it is.

So how is it that the conductor is repelled by both the north field and the south field ?.

The last 2 pictures make more sense.


Brad

Case 1:
You can see that the field lines generated by the wires near the magnet are about the same as if the wire were a small N'-S' magnet parallel to the real magnet and oriented in the same direction. It is therefore repelled.

Case 2
The equivalent small magnet that the wire simulates has now is S'-N' poles reversed, it is attracted.

Woops! I got the direction wrong (used Brad's image).  Each line crossing obeys Fleming's LH rule so here is the revised image.  You can see it is one half of Brad's figure of eight field.
Smudge

Quote from: Smudge on 2019.06.11, 17:12:58
Woops! I got the direction wrong (used Brad's image).  Each line crossing obeys Fleming's LH rule so here is the revised image.  You can see it is one half of Brad's figure of eight field.
Smudge

;)    Yup.  When you have a vision of the fields, its pretty simple. ^-^

Mags

One thing about this that has my mind going is when doing the loop around is the outward push on the wire while it is being drawn around from one side of the pole the other. Not that I dont understand it, just how can we maybe use this and could it hold any possible advantages. When we have like I described earlier, a window motor type rotor to hold the wire in the circular path, these other inward and outward forces are not apparent as when the wire is loose.  Like I said before, it is still interesting to see what we wouldnt have before this. O0

Mags

Apply ac to the wire.  Remember those light bulbs back in the 70s and 80s with a filament in the shape of a candle flame outline with a little magnet mounted between the 2 posts that the filament is connected to?  ^-^

Mags

Quote from: Smudge on 2019.06.11, 14:53:40
Brad, Itsu, PM, F6, Guys,
You are comparing apples to oranges.  Field lines are lines of force, and here we have two different lines of force.  First off the B field lines show the direction of a force that would apply to an isolated magnetic N pole.  Such an isolated N pole would be repelled from the N pole of the magnet and would travel along one of those field lines to reach the S pole of the magnet.  We don't have isolated poles but you can do an experiment that will show this effect.  Make a long thin magnet by magnetizing a length of rigid piano wire, then suspend this from a ceiling so that it hangs down as a pendulum with the S pole at the suspension point, a pendulum that can follow the surface of a cone.  Then place a small magnet directly below it to see the N pole of the pendulum repelled from the N pole of the magnet.  You will see that N pole follow a curved trajectory as depicted in the rather poor image attached.   That is showing you the B field line.  It is a line of force and if you integrate that line from The N pole of the magnet to the S pole you get the energy (force times distance) or work done.  So from that starting position there is that potential energy available.  It is not just some vague potential, it is potential energy.  If you started somewhere else on the force line you would get less potential energy.  If you wanted to you could draw other "field" lines like lines of equipotential energy.  Iron filings are magnetized by the B field to become tiny magnets and they do link up to form strings that follow those B lines.

Brad's figure of eight field lines are a different field.  They are not the force on an isolated N pole, they are the force on a line element of current.  Brad is correct to see them as field lines, and to call them magnetic field lines as clearly they come from magnetism.  They need a different terminology.  We already have magnetic A fields, magnetic B fields and magnetic H fields.  Perhaps we should call them F8 fields because of that figure of 8.  Or maybe we could use FLH fields as they obey Fleming's LH rule.  Or why not just magnetic F field?  Since by Fleming's LH rule the force is always at right angles to the B field, you can easily plot the F field lines as shown in the second image.   And being lines of force there is potential energy available so you could also plot lines of equipotential energy of you wanted to.
Smudge

Perhaps that helps explain those !other! experiments ?.


Brad

Quote from: TinMan on 2019.06.12, 05:16:53
Perhaps that helps explain those !other! experiments ?.
Brad

With an open magnet near the slip-ring there is always the possibility that we are measuring vXB induction (Fleming's RH rule), as F6 tried to point out.   The first image below shows that induction occurring at the brush contacts with voltage induced across the slip-ring width, not around its circumference.  Then it might be possible that we see a voltage at the brush contacts if they are off to one side (as shown).  Scientists who are brainwashed into believing that vXB is the only form of induction and that a non curl A field (no B field) can't induce will always put up this argument.

To dispel that argument we really need to eliminate the external B field, i.e it should be contained within a closed magnetic circuit.  The second image shows such an experiment using a PM where the magnetic flux is shunted through soft perm material.  The third image is the real killer, that uses a ring core with a toroidal winding on it.  If that produces an induced DC voltage across the slip-ring when there is DC current in the coil, or an AC voltage when there is AC current in the coil, then we have definite proof that the non-curl A field can be detected and used.

Smudge

Hi Smudge,

If -E=(v×∇).A, the simplest proof is a Hall effect sensor submitted to A where B=0. Only the Lorentz force is the presumed cause of their functioning.
It's an easy test to do, we avoid mechanics. These sensors output from 1 to 5 mV/Gauss, so they are quite sensitive. If they react with A instead of B, we win.
I don't have much time now and I have to order this cheap electronic component, but I will experiment it.

Quote from: F6FLT on 2019.06.12, 16:35:54
Hi Smudge,
If -E=(v×∇).A, the simplest proof is a Hall effect sensor submitted to A where B=0. Only the Lorentz force

My book tells me that triple product (v×∇).A is a scalar so I don't see where you are coming from.
Edit>  If you mistakenly put (v×∇).A when you meant to put the convective term (v.∇)A that I have banged on about then I still don't see why you expect the Hall sensor might detect the A field.

Quote from: Smudge on 2019.06.12, 16:55:39
... If you mistakenly put (v×∇).A when you meant to put the convective term (v.∇)A that I have banged on about then I still don't see why you expect the Hall sensor might detect the A field.

You're right. It's (v.∇)A that I wanted to talk about. By dint of jumping from the vector product of Lorentz force to the vector potential, we end up mixing the formulas.  :-[
However, the idea remains a good one.

Unlike a disc where the speed of electrons is obtained by its rotation, in a Hall effect sensor it is an input current that makes the speed.
But as in a Faraday disc where a transverse voltage (radial) is measured, it is this transverse voltage that the component outputs.
The transverse voltage is due to the Lorentz force according to the classical explanation, and to the force related to the spatial gradient of A in the explanation by the vector potential.

This spatial gradient of A must therefore produce the same effect as the Lorentz force where it exists and B does not, and consequently influence the Hall effect sensor outside of a magnetic field, which is to be verified.

Quote from: F6FLT on 2019.06.13, 06:53:13
You're right. It's (v.∇)A that I wanted to talk about. By dint of jumping from the vector product of Lorentz force to the vector potential, we end up mixing the formulas.  :-[
However, the idea remains a good one.

Unlike a disc where the speed of electrons is obtained by its rotation, in a Hall effect sensor it is an input current that makes the speed.
But as in a Faraday disc where a transverse voltage (radial) is measured, it is this transverse voltage that the component outputs.
The transverse voltage is due to the Lorentz force according to the classical explanation, and to the force related to the spatial gradient of A in the explanation by the vector potential.

This spatial gradient of A must therefore produce the same effect as the Lorentz force where it exists and B does not, and consequently influence the Hall effect sensor outside of a magnetic field, which is to be verified.

OK but I must point out that a primary feature of (v.del)A is its longitudinal components that are absent in vXB,  And the longitudinal induction will not create the Hall effect.  Still worth doing though.
Smudge

Quote from: Smudge on 2019.06.13, 18:05:13
OK but I must point out that a primary feature of (v.del)A is its longitudinal components that are absent in vXB,  And the longitudinal induction will not create the Hall effect.  Still worth doing though.
Smudge

So what is for you the equivalent of vXB for A?

For me: there is no longitudinal induction in the Faraday disk, only a transverse one due to the relativistic effect of length contraction, because there is no length contraction when the observed length is colinear to the speed vector. The equivalent of vXB for A comes from the Lorentz transformation of the electromagnetic 4-vector which combines A and the scalar potential.
It is named A^µ. A^µ=(φ/c,Ax,Ay,Az). Each coordinate of A^µ are function of the space-time position (ct,x,y,z) we are looking at. A^µ is transformed into A'^µ as follows:

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

At the beginning, we have no scalar potential in the referential at rest: A^µ=(0,Ax,Ay,Az). Suppose that A is only along x, so Ay=Az=0. You obtain φ/c=-γβAx and A'x=γAx. We are interested here in φ/c which is the scalar potential that appears in the referential of the moving charge from the relativistic effect.
When the velocity v is different at different places in space, as is the case along the radius of a rotating Faraday disc, then φ/c=-γβAx changes with the speed which is included in β=v/c and increases from the center of the disc to the rim. Therefore we obtain a potential difference along the radius, which corresponds to the electric field E=vXB that you see with the Lorentz force.

About the longitudinal induction:
in absence of a scalar potential, a longitudinal induction (say along x) means a longitudinal potential difference viewed by the moving observer (the charge) 𝝯φ/c = γAx1-γAx2≠0, or γAx(ct1,x1,0,0)-γAx(ct2,x2,0,0)≠0, i.e. that Ax has not the same value at both positions, which implies either a time dependent variation of A (classical induction), or a longitudinal spatial gradient of A along x, or a not constant speed of the observer, the charge (this last case is to be verified with GR, as the observer's referential is no more inertial).

Quote from: F6FLT on 2019.06.14, 09:46:19
So what is for you the equivalent of vXB for A?

I am not well versed in using the determinant form of vectors to it will take me some time to convert your 4 vector into something meaningful for me.  My version is in equation A2 of my paper in Reply #51 of this thread.

Quote
About the longitudinal induction:
in absence of a scalar potential, a longitudinal induction (say along x) means a longitudinal potential difference viewed by the moving observer (the charge) γAx1-γAx2≠0, or γAx(ct1,x1,0,0)-γAx(ct2,x2,0,0)≠0, i.e. that Ax has not the same value at both positions, which implies either a time dependent variation of A (classical induction), or a longitudinal gradient of A along x, or a not constant speed of the observer (the charge).

I don't disagree with that and the non curl A fields can have just that longitudinal gradient of A along x.  Those longitudinal gradients appear in the convective derivative (v.del)A.  But they don't appear in the (vXB) term.  My take on this is that the curl function creates a field where you can't get a longitudinal effect.  So they don't appear in the Faraday disc.  But in a non-curl A field they do exist.  And it is up to us to find that effect.
Smudge

 

Quote from: Smudge on 2019.06.14, 10:41:47
I am not well versed in using the determinant form of vectors to it will take me some time to convert your 4 vector into something meaningful for me.  My version is in equation A2 of my paper in Reply #51 of this thread.I don't disagree with that and the non curl A fields can have just that longitudinal gradient of A along x.  Those longitudinal gradients appear in the convective derivative (v.del)A.  But they don't appear in the (vXB) term.  My take on this is that the curl function creates a field where you can't get a longitudinal effect.  So they don't appear in the Faraday disc.  But in a non-curl A field they do exist.  And it is up to us to find that effect.
Smudge

The advantage of using relativity in electromagnetism is that it is complete, because we are really in a 4D universe, and contrary to what we would think, it also makes it easier to imagine qualitative effects.

By using 4-vectors, no more approximations or oversights are made.
For example, a movement relative to the magnetic vector potential will show you a scalar (electric) potential. But also you will no longer see the same vector potential.
However, the electrical potential, as well as the magnetic vector potential, change in a covariant way, hence the interest of treating them together in a formalism that supports it, rather than separately.

For example, we know that the field is derived from a potential. In a 4D space, a vector, like the potential vector, has 4 coordinates.
If we want to move to the fields, we have 16 gradients, one for each difference between each of the 4 coordinates of the 4-vector potential at one point in space-time and each of the 4 coordinates of the 4-vector potential at another point in space-time.
So we get a 4x4 matrix, and this one really gives us _all_ the electromagnetic information at one point in space-time, it's the electromagnetic tensor.

It is noted F^µν=∂^µA^ν - ∂^νA^µ, according to Einstein's convention (https://en.wikipedia.org/wiki/Einstein_notation). And this gives us the fields we are familiar with:
F^µν=
| 0      -Ex/c  -Ey/c  -Ez/c |
| Ex/c   0      -Bz      By    |
| Ey/c   Bz     0       -Bx    |
| Ez/c  -By     Bx      0      |

If we see so many contradictory opinions in electromagnetism, it is because traditional formalism is insufficient.
Special relativity is not new to me, but its use in electromagnetism, yes. I am trying to familiarize myself, and I regret that electromagnetism was not taught to me on this basis, it is really elegant, complete and uncompromising.

For anyone interested in here is my latest paper looking into motional induction from the magnetic vector potential A field.  By taking account of both the change of vector amplitude and the change of vector orientation with distance it is possible to obtain classical E = (v × B) motional induction directly from the time-rate-of-change of the magnetic vector potential A. When this procedure is applied to a non-curl A field where B is zero it predicts both a longitudinal {equations 8 or 11} and a transverse {equations 10 or 12} induced force on moving charge.
Enjoy!

Smudge

Hi Smudge

I find your paper fascinating, can I pass a question by you as what you think would be the outcome in the following:-

A coax cable which has the center core of iron with a coating of copper, a dielectric insulator and then a copper screen.
The coax is bent to form a loop or multiple loops (coil) where the ends are magnetically and electrically looped together.
So we now have a ferromagnetic core and an inductor/capacitor with an insulating dielectric between the inner and outer copper conductors.
Around the now toroidal looking loops we wind a secondary coil or coils (our output) at 90º.
To the two copper conductors of the coax, we supply an alternating HF current which is at the natural resonant frequency of the capacitive/inductive coax coil/ toroid.

I am thinking of the magnetic fields interreacting with the capacitive charge (a possible magnetic or electron accelerator which would induce charge into the secondary coil).

Thanks in advance

Regards

Mike 8)
ADDED
PS the whole has a ferromagnetic core, the inner conductor

Quote from: Centraflow on 2019.07.28, 10:33:04
Hi Smudge

I find your paper fascinating, can I pass a question by you as what you think would be the outcome in the following:-

A coax cable which has the center core of iron with a coating of copper, a dielectric insulator and then a copper screen.
The coax is bent to form a loop or multiple loops (coil) where the ends are magnetically and electrically looped together.

Not sure what you mean here.  Are the inner and outer ends connected so we have a shorted turn of it iron wire that is surrounded by a shorted turn of the coax outer?  Please clarify.

Smudge

Quote from: Smudge on 2019.07.28, 16:29:16
Not sure what you mean here.  Are the inner and outer ends connected so we have a shorted turn of it iron wire that is surrounded by a shorted turn of the coax outer?  Please clarify.

Smudge

Sorry I was not clear.

Take a piece of coax which has the center core made of iron but coated with copper (that is then a ferromagnetic core and a conductor. Make multiple turns and connect the center core ends together, then the screen to screen ends together so as both inner and outer wires are independent. We would now have both a capacitor and a coil which has a magnetic core.

Hope that explains a bit better :)

Regards

Mike 8)

OK Mike,
I get it now.  The coax outer will be a shorted turn to the secondary coil wound over it.  Thus the field from the secondary induces eddy currents around the coax and those eddies will cancel out the field thus acting as a magnetic shield preventing the inner core from being magnetized.  Only at very low frequency will that inner core work as an inductor to get any excitation.  And then it can't be resonant.
Smudge

Quote from: Smudge on 2019.07.29, 08:40:08
OK Mike,
I get it now.  The coax outer will be a shorted turn to the secondary coil wound over it.  Thus the field from the secondary induces eddy currents around the coax and those eddies will cancel out the field thus acting as a magnetic shield preventing the inner core from being magnetized.  Only at very low frequency will that inner core work as an inductor to get any excitation.  And then it can't be resonant.
Smudge

Hi Smudge,

ok, both the inner (copper coating on the iron) is a shorted turn and the outer is not shorted turn so they are like the two plates of a capacitor, in my case 1.22nF, and 0.415mH inductance on the outer screen if not shorted, the inner of course is a shorted turn. The resonant frequency is 223.7kHz.

Around this wound at 90º is the secondary, question is will the secondary see the primary? will the iron center core have eddy currents? what will happen to the capacitive charge in relation to the magnetic field of the core?

Can you see what I am getting at?

Regards

Mike 8)

Quote from: Centraflow on 2019.07.30, 11:27:58

Can you see what I am getting at?

Regards

Mike 8)

Not really.  Can you show a sketch or a picture of the set up please?  Just the single loop of coax would do to show me where the input is fed.

Smudge

 

Smudge

see your PM's

Regards

Mike 8)

Here are a few images showing the A field external to a core.  If the flux in the core is varying with time, for any closed circuit that encloses the core we of course get an E field driving current around that circuit.  For a closed circuit that does not enclose the core there are still E fields, but they do not drive current around the circuit because they cancel out wrt to voltage induction.  But they can polarize the circuit, produce surface charges at different parts of the circuit.  Could this be used to advantage?

For constant flux the A field doesn't vary with time, but electrons moving through that non-linear A field can "feel" a time changing A and therefore an E field.  That E field at any point is proportional to the electron velocity.  In any closed circuit, if electrons travel at constant velocity the sum of any induced voltage is zero, there is no motion-induced voltage to maintain any current.  But there are some parts of the loop where the motion-induced E field supports the current and other parts where it opposes the current.  Again this will create surface charge on the conductor.

In the Marinov generator the motion induced voltage across the slip-ring, coming from relatively high slip-ring velocity, is not negated by the motion-induced voltage from the rest of the closed circuit, because there the velocity is only trivial drift velocity.  So the generator can drive current through a load.  And since it doesn't work in reverse as an electric motor it should be inherently OU.

Smudge

A revision of Maxwell Theory, supporting the Marinov's Generator.
Published by Mario J. Pinheiro from the University of Lisbon.

I am re-invigorating this topic in the hope that someone might carry out the experiment.  Tinman (Brad) posted a budget version on this thread in 2018, see first image below.  I posted another image on another thread in June 2019 where Brad said he would have a go, image 2 below.  Brad is a very busy man so I'm hoping someone else will take up the challenge.

Of all the OU devices I have worked on this one promises to come true since we can see where the energy comes from.  The version presented here can have DC current in the toroid thus becoming a replication of Marinov's version using magnets.  The advantage of the toroid is that we can drive with AC and that allows us to use a transformer to step up the small output voltage to more meaningful levels.  If it is shown to work it will answer a fundamental principal that is still the subject of much controversy in the scientific world, can you extract energy from a magnetic vector potential A field where there is zero magnetic B field, where curl A = zero.

Smudge

Just a note
Did start mirror topic
Here https://overunity.com/18889/builders-board-discussion-for-smudges-marinov-experiments/msg558803/#msg558803

(Smudge is moderator there also ( his section there)

With gratitude
Chet

Smudge
That is an interesting concept and I have done quite a few experiments which relate to this subject.

1) A magnet made of conductive material spun on it's N-S axis will develop an induced voltage from the center to the perimeter not unlike a homopolar generator. This is true because the magnetic field exists in the space within and around the magnet itself and does not rotate with the magnet. I developed a magnetic bearing stabilizer based on this concept. I used an aluminum plate adjacent to the magnet and no eddy currents were induced by the spinning magnet until it moved off center in which case eddy currents were generated centering/stabilizing the spinning magnet. Obviously my stabilizer would not work and induce a large drag force if the magnetic field rotated with it's source ergo it does not.

2) A large turn coil like a Tesla coil spun on the horizonal axis with brushes on each end will also develop a voltage across it and a current in a closed circuit. This is true because the coil is in an electric field gradient, the Earth has a negative charge and the atmosphere a positive charge increasing with elevation. As such the free electrons are always repelled from the Earth and attracted to the upper atmosphere in a conductor. When we spin the coil the free electrons want to rise from both sides of the coil upwards however the conductor in spinning. Thus the free electrons trapped at the top of the coil find themselves in a conductor with a corkscrew motion and with each turn they move horizontally. The action is no different than a spiral pump shown below only the fluid is electrons instead of water.

This is the beauty of the scientific method in developing a basic fundamental theory of how stuff works for ourselves then proving it by doing relatively inexpensive experiments and gaining a wealth of knowledge in the process.

I use a very different approach than most and my thoughts revolve around forces and motion on whatever level they may occur. I try to avoid math and specific terminology as well as the usual trappings of circular logic supposing that just because something is considered normal here it is considered normal everywhere. The universe would seem to be a very big place with many different perspectives and I think we must have concepts common to everyone relating to nature.

Regards
AC

I assume that the reason homopolar generators and motors attract little attention is because they operate in an unusual regime (low voltage, high current) where losses like brush friction and resistance make them unviable commercially.  However these objections can be overcome by using rolling contacts.  Of particular interest is the possibility of driving the Marinov generator with a classical homopolar motor (like the Faraday Disc).  The DC output of the Marinov generator could drive the homopolar motor (compatibility wrt low voltage, high current) which in turn mechanically drives the generator.  The OU potential of the Marinov generator offers the potential for the system to free run and at the same time deliver excess shaft energy to something else.  The electrical connection between the two would be via the axles of the rolling brushes, which can carry the high current needed.  Opportunities exist here for some innovative designs.  Unlike other free running magnetic motors, the excess energy can be accounted for via a visible link to the quantum domain.

Quote from: Allcanadian on 2021.06.24, 01:23:58
Smudge
That is an interesting concept and I have done quite a few experiments which relate to this subject.

1) A magnet made of conductive material spun on it's N-S axis will develop an induced voltage from the center to the perimeter not unlike a homopolar generator. This is true because the magnetic field exists in the space within and around the magnet itself and does not rotate with the magnet. I developed a magnetic bearing stabilizer based on this concept. I used an aluminum plate adjacent to the magnet and no eddy currents were induced by the spinning magnet until it moved off center in which case eddy currents were generated centering/stabilizing the spinning magnet. Obviously my stabilizer would not work and induce a large drag force if the magnetic field rotated with it's source ergo it does not.

Interesting demonstration that the magnetic field does not spin.  For myself I concluded that the trivial spin speeds we produce just can't compare with the spin speeds of orbital electron or the electrons themselves, hence our applied spin does nothing to the magnetic field.

Quote
2) A large turn coil like a Tesla coil spun on the horizonal axis with brushes on each end will also develop a voltage across it and a current in a closed circuit. This is true because the coil is in an electric field gradient, the Earth has a negative charge and the atmosphere a positive charge increasing with elevation. As such the free electrons are always repelled from the Earth and attracted to the upper atmosphere in a conductor. When we spin the coil the free electrons want to rise from both sides of the coil upwards however the conductor in spinning. Thus the free electrons trapped at the top of the coil find themselves in a conductor with a corkscrew motion and with each turn they move horizontally. The action is no different than a spiral pump shown below only the fluid is electrons instead of water.

The earth's E field is little used.  At the surface it is something like 300V/m on a clear day (or night) but it does change when there is cloud cover, and that change can be a reversal of the field as a highly charged cloud moves overhead.  I did offer the possibility that dowsing uses the earth's E field.  The Y shaped stick that some dowsers use concentrates the E field emanating from the human-body shaped extension to the earth's surface, (it is well known that surface charge concentrates at sharp points).  The stick is held in a position of unstable equilibrium where the slightest change of the E force on the charged tip triggers its movement.  That change comes from the presence of the underground water, its higher electrical conductivity wrt the soil influences the surface E field.  A sensitive E field detector could detect the same thing.

Smudge

Smudge

Was uncertain as to the exact concept of the Marinov Motor/Generator until I found this video... https://www.youtube.com/watch?v=iTuDrJVzCpo

It would seem to be a variation of the homopolar motor/generator in my opinion. It's not common knowledge but supposedly closed magnetic paths are seldom if ever closed. I have measured external magnetic fields in all closed transformers well below saturation and closed circuits using magnets. I found this to be true because there is always leakage where laminations or magnets interface or imperfections in the domains. There is also the fact that one condition(a North pole) cannot transition to an opposite condition(a South pole) without moving through a zero plane or neither condition in the process. So any resistance/reluctance in a system must produce discontinuity on some level in principal.

QuoteInteresting demonstration that the magnetic field does not spin.  For myself I concluded that the trivial spin speeds we produce just can't compare with the spin speeds of orbital electron or the electrons themselves, hence our applied spin does nothing to the magnetic field.

It is no coincidence that the Magnetic, Electric, Gravic, Inertial fields and light do not spin with there source. No wave property does spin with it's source which gives us some insight into it's nature.

I find most peoples thinking on this subject to be bizarre to be honest. Now suppose I had a cam shaped cylinder spinning in water producing a periodic wave like disturbance due to the cam displacing a volume of water. Should I suppose that the displaced water or wave it produces must spin with the source?. It seems absurd because the displacement is always at right angles to the source of the displacement.

QuoteThe earth's E field is little used.  At the surface it is something like 300V/m on a clear day (or night) but it does change when there is cloud cover, and that change can be a reversal of the field as a highly charged cloud moves overhead.

Many people know that Earth's E field is around 300V/m but few have considered that 300V/m is equal to 3V/cm or 0.3V/mm. So while the difference in voltage may appear insignificant on smaller scales it does not change the fact that Power is equal to the difference in potential times the rate of change or current. So if we could cycle a small difference in potential at a very high rate of change we still get tangible power generation.

As Einstein said... nothing happens until something moves however what he didn't say is that motion is energy. Thus the rate of change or motion must always be an integral part of our consideration of the Energy something has. The absolute "rate of change" has many facets to consider which also relates to the total energy of something within a system.

Regards
AC

Quote from: Allcanadian on 2021.06.25, 00:45:53
Smudge

Was uncertain as to the exact concept of the Marinov Motor/Generator until I found this video... https://www.youtube.com/watch?v=iTuDrJVzCpo
It would seem to be a variation of the homopolar motor/generator in my opinion.

That device is not true to the Marinov device, the slip-rings are in the wrong place.  And you may be right, that might be using classical flux cutting to produce the motion.

QuoteIt's not common knowledge but supposedly closed magnetic paths are seldom if ever closed. I have measured external magnetic fields in all closed transformers well below saturation and closed circuits using magnets. I found this to be true because there is always leakage where laminations or magnets interface or imperfections in the domains. There is also the fact that one condition(a North pole) cannot transition to an opposite condition(a South pole) without moving through a zero plane or neither condition in the process. So any resistance/reluctance in a system must produce discontinuity on some level in principal.

I spoke with a professor at Delft University and he showed me what he thought was a great discovery, a leakage field in a transformer that had primary and secondary coils on opposite sides of the closed core magnetic path.  I had to explain that this was not something new, Lenz's Law produces a secondary field that counters the primary field and must drive flux outside the core if the secondary is not wound over the primary.

I don't think the Marinov generator is a classical homopolar device, classical physics does not allow induction along a moving conductor when the velocity of the conductor is along its length.  All classical generators use conductors that have their velocity vector at right angles to the conductor line-element.  This slip-ring generator induces voltage along the conductor while the velocity is also along the conductor.  That is something new.

Smudge