The previous negative result prompted me to revise the theory. The reasoning up to now was done in the framework of Galilean kinematics, but electromagnetism is by nature relativistic, and only its framework allows a correct analysis.
I spent many painful hours working on it, even though the method was quite simple, but I still had to find it. I think I found the exact maths for a gradient of the vector potential through special relativity.
As a reminder:
We know that an electric field E is created by the time variation ∂A/∂t (which corresponds to the magnetic induction) and by a potential difference ∇φ.
The general case is E=-∂A/∂t-∇φ, where A is the vector potential and φ the scalar potential.
We assume a volume where φ=0 (no electric field deriving from a potential) and where A is oriented along the x axis, and has a spatial gradient ∇Ax. If a charge is moving at speed v along x, it should see a variation of A related to its position, hence an electric field by the virtue that E=-∂A/∂t=-∂A/∂x.∂x/∂t =-v.∂A/∂x.
This is the experiment that was proposed here and whose theory we want to verify.
In the framework of special relativity, A is a 4-vector to which the Lorentz transforms apply as a 4x4 matrix:
│φ'/c│ │ γ -γβ 0 0│ │φ/c│
│A'x │ = │-γβ γ 0 0 │ │Ax │
│A'y │ │ 0 0 1 0│ │Ay │
│A'z │ │ 0 0 0 1│ │Az │
φ is the scalar potential, Ax, Ay, Az the values of the magnetic vector potential A on the 3 axes, β = v/c with v the velocity of the charge, and γ = 1/√(1-β²) is the Lorentz factor. A and φ are seen by the observer, A' and φ' by the charge.
In our case φ, Ay and Az are zero.
We simply apply the matrix calculation:
A'x = -γ.β.φ/c + γ.Ax = γ.Ax because φ=0
φ'/c = γ.φ/c - γ.β.Ax = - γ.β.Ax because φ=0
hence φ' = - γ.β.c.Ax = -γ.v.Ax
and ∇φ' = -γ.v.∇Ax
=>
E' =-∂A'/∂t-∇φ' = -γ.∂Ax/∂t + γ.v.∇Ax = -γ.v.∂Ax/∂x + γ.v.∇Ax = 0 In other words: the electric field produced by the temporal variation that the charge sees due to the spatial gradient of the vector potential, is exactly compensated by the field from the gradient of a scalar potential that appears due to its movement!
The worst thing is that I had noticed this scalar potential 4 years ago and then completely forgotten about it
. See
https://www.overunityresearch.com/index.php?topic=2470.msg74654#msg74654 where I pointed out that
"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".The negative result of the experiment is therefore normal. It will be necessary to be more subtle to exploit the basic idea, for example by playing on the 3 space coordinates of A and perhaps even adding a scalar potential or making the speed of the charge variable...
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Topic: New generator from a spatial gradient of the vector potential and current (Read 7022 times)
