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Author Topic: Marinov Generator  (Read 22549 times)
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... 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.



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"Chance favours only the prepared mind."  Louis Pasteur
   
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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.
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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 Aµ. Aµ=(φ/c,Ax,Ay,Az). Each coordinate of Aµ are function of the space-time position (ct,x,y,z) we are looking at. Aµ 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: Aµ=(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-γAx20, 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).

« Last Edit: 2019-06-14, 11:44:27 by F6FLT »


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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-γAx20, 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
   
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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 Fµν=∂µAν - ∂νAµ, according to Einstein's convention. And this gives us the fields we are familiar with:
Fµν=
| 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.



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"Chance favours only the prepared mind."  Louis Pasteur
   
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