[Physics] Mathematical proof Maxwell's equations are incorrect?

Arend Lammertink lamare at gmail.com
Thu Apr 23 10:48:00 CEST 2020


Hi all, reworded the crux of the argument in just a few lines of math:

-:-
When we start with Laplace's (or Poisson's) equation, we have a 2nd
order differential equation in 3D. It doesn't matter whether one
wishes to use Laplace's equation or the more general Poission
equation, since in both of these the Laplacian represents the flux
density of the gradient flow of a function.

In one dimension, the Laplacian simply is βˆ‚Β²/βˆ‚xΒ² , i.e., the curvature.

In other words: the Laplacian IS the second order spatial derivative
of ANY given vector funtion 𝐅, the 3D curvature if you will, and is
given by the identity:

 βˆ‡Β²π…= βˆ‡(βˆ‡Β·π…) - βˆ‡Γ—(βˆ‡Γ—π…)

The terms in this identity can be written out as follows:

 𝐀=βˆ‡Γ—π…
 Ξ¦= βˆ‡β‹…π…
 𝐁=βˆ‡Γ—π€=βˆ‡Γ—(βˆ‡Γ—π…)
 π—˜=βˆ’βˆ‡Ξ¦= βˆ’βˆ‡(βˆ‡β‹…π…)

And because of vector identities, one can also write:

 βˆ‡Γ—π—˜= 0
 βˆ‡β‹…π= 0

So, at the very least we can establish that because in Maxwell's
equations we have:

 βˆ‡Γ—π—˜= -βˆ‚π/βˆ‚t,

which is not 0, Maxwell's equations are NOT the second order spatial
derivative of ANY function 𝐅.

Or, to put it the other way around:

Because βˆ‡Γ—π—˜is not 0 in Maxwell's equations, no function 𝐅exists for
which Maxwell's equations are the 2nd order spatial derivative.

 -:-

Best regards,

Arend.




On Thu, Apr 23, 2020 at 8:09 AM Arend Lammertink <lamare at gmail.com> wrote:
>
> Dear List members,
>
> I have been studying Tesla for quite some time now and became
> convinced longitudinal waves exist and that they propagate faster than
> light. For quite some time, I have been working on the theory, which
> culminated in the attached draft paper on revision of Maxwell's
> equations. During the past week, I had a discussion about this on the
> "Theoretical Physics" LinkedIn group, which made me realise how
> important the vector Laplace equation is and believe I now have the
> mathematical proof that Maxwell's equations are incorrect. This is the
> short version of the argument:
>
> -:-
> "The Laplace operator is not some sacred physical law of the universe,
> it is a mathematical relation".
>
> Yes, it's a relation of which the correctness is pretty much
> undisputable, like 1+1=2.
>
> Equate this equation to zero and one obtains the 3D Laplace equation
> of which the solutions are the harmonic functions, which (when worked
> out) describe all possible (harmonic) wave phenomena in 3D:
>
>  βˆ‡Β²π…= βˆ‡(βˆ‡Β·π…) - βˆ‡Γ—(βˆ‡Γ—π…) = 0.
>
> This can be re-written as:
>
>  -βˆ‡Β²π…= - βˆ‡(βˆ‡Β·π…) + βˆ‡Γ—(βˆ‡Γ—π…) = 0.
>
> Then, the terms in this equation can be written out as follows:
>
>  𝐀= βˆ‡Γ—π…
>  Ξ¦= βˆ‡β‹…π…
>  𝐁= βˆ‡Γ—π€= βˆ‡Γ—(βˆ‡Γ—π…)
>  π—˜=βˆ’βˆ‡Ξ¦= βˆ’βˆ‡(βˆ‡β‹…π…)
>
> And because of vector identities, one can also write:
>
>  βˆ‡Γ—π—˜= 0
>  βˆ‡β‹…π= 0
>
> So, any given vector field 𝐅 can be decomposed like this into a
> rotation free component π—˜ and a divergence free component 𝐁.
>
> There is no argument this is mathematically consistent, nor that the
> solutions to the equation -βˆ‡Β²π…= 0 are the harmonic wave functions in
> 3D.
>
> Now compare this to Maxwell's:
>
>  π—˜= βˆ’βˆ‡Ξ¦βˆ’ βˆ‚π€/βˆ‚t
>
> Take the rotation at both sides of the equation and we obtain the
> Maxwell-Faraday equation:
>
>  βˆ‡Γ—π—˜= - βˆ‚π/βˆ‚t
>
> WP: "Faraday's law of induction (briefly, Faraday's law) is a basic
> law of electromagnetism predicting how a magnetic field will interact
> with an electric circuit to produce an electromotive force (EMF)β€”a
> phenomenon known as electromagnetic induction."
>
> This is a circuit law, which predicts how a magnetic field will
> interact with electrons moving trough a wire. Since this involves
> moving charge carriers, which are particles, it is illogical to
> introduce this law at the medium/field modelling level. Because of the
> wave-particle duality principle, it is known that particles are
> manifestations of the EM field. So, by including this law in the
> medium/field model one introduces circular logic.
>
> Not only that, it breaks the fundamental separation of the fields into
> a divergence free component and a rotation free component.
>
> As is well known, this model eventually leads to two mutually
> exclusive theories, which cannot both be correct.
>
> In other words: what you are doing by introducing Faraday's law at
> this level in the model is you are insisting 1+1 is not 2, but
> something else.
>
> And you end up with 150+ years of trying to find additional equations
> to straighten things out, but the bottom line is: 1+1=2, NOT something
> else
>
> [...]
>
> "How does it break "the fundamental separation of the fields into a
> divergence free component and a rotation free component."? "
>
> As shown, the 3D vector Laplace equation defines two components, one
> of which is divergence free and one of which is rotation free.
>
> Since the 3D vector Laplace equaton is nothing but a 3D generalization
> of the lower dimensional Laplace equation and results in harmonic
> solutions, which is all well established undisputable math, it follows
> that the decomposition into a divergence free component and a rotation
> free component is fundamental and is therefore the only correct way to
> derive wave functions in 3D for any given vector field.
>
> There is no argument that with equating the rotation of the rotation
> free component π—˜ to the time derivative of the divergence free (and
> therefore rotational) component 𝐁 by Maxwell results in π—˜ remaining
> to be rotation free and therefore such breaks said fundamental
> separation of said components.
> -:-
>
> I have some rewriting to do of the article, because I now realize it's
> perfectly O.K. to have the primary field, which I denoted [V], as the
> null vector field, since in the Laplace equation the right side of the
> equation is also zero, so we don't have to resort to discrete math.
> So, for the time being, I included part of the discussion on LinkedIn,
> which I think you'll find interesting.
>
> In short: I believe to have found the foundation for that Theory of
> Everything scientists have been looking for for a very long time.
>
> I would love to hear your opinion about this.
>
> Best regards,
>
> Arend.



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