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Author Topic: Marinov Generator  (Read 20018 times)
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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.




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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=delA(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.
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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
   
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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.



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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.
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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.


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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.

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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.

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.

Quote
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.

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.

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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.



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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.
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and 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.
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I 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.
Quote
I 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.
Quote
I 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.

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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" 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.



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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).


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tExB=qr
See the attached document:

(Sorry if I posted this before, I have a poor memory.)
   
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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.
« Last Edit: 2019-05-20, 19:46:51 by Smudge »
   
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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. :)

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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.

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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.


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"Chance favours only the prepared mind."  Louis Pasteur
   
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