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

[Arend Lammertink]

Hi Hans and General,

I know you have interesting math which is way above my head, but the
problem is not how to describe fields and the dynamics of the aether.
The problem is Maxwell didn't adhere to the Laplace equation and that
is what caused not only relativity, but QFT as well. The wave equation
that comes out of Maxwell is not invariant to the Galilean transform,
hence the invention of the Lorentz transform, and because he
introduced Faraday's law, a circuit law, at the wrong level to the
model, his model is not defined along the Laplace operator, which is
why an unwarranted gauge freedom entered into the model, hence the
invention of gauge theory.

So, we ended up with 150+ years of trying to find additional equations
in order to straighten things out, eventually culminating in two
pretty much "sacred" theories that are mutually exclusive; they cannot
both be correct.

So, the question here is whether or not I made a mistake somewhere and
whether or not my argument counts as mathematical proof for the
incorrectness of Maxwell's equations. This is the argument as I have
it now:

-:-
In potential theory, the study of harmonic function, the Laplace
equation is very important, amongst other with regards to
consideration of the symmetries of the Laplace equation. The
symmetries of the n-dimensional Laplace equation are exactly the
conformal symmetries of the n-dimensional Euclidean space, which has
several implications. One can systematically obtain the solutions of
the Laplace equation which arise from separation of variables such as
spherical harmonic solutions and Fourier series. By taking linear
superpositions of these solutions, one can produce large classes of
harmonic functions which can be shown to be dense in the space of all
harmonic functions under suitable topologies.

The Laplace equation as well as the more general Poisson equation are
2nd order differential equations, in both of which the Laplacian
represents the flux density of the gradient flow of a function. In one
dimension, the Laplacian simply is ∂²/∂x² , representing the curvature
of a given function f. For scalar functions in 3D, the Laplacian is a
common generalization of the second derivative and is the differential
operator defined by:

∇²f = ∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z².

The Laplacian of a scalar function is equal to the divergence of the
gradient and the trace of the Hessian matrix. The vector Laplacian is
a further generalization in three dimensions and defines the second
order spatial derivative of any given vector function 𝐅, the 3D
curvature if you will, and is given by the identity:

∇²𝐅= ∇(∇·𝐅) - ∇×(∇×𝐅)

Whereas the scalar Laplacian applies to a scalar field and returns a
scalar quantity, the vector Laplacian applies to a vector field,
returning a vector quantity. When computed in orthonormal Cartesian
coordinates, the returned vector field is equal to the vector field of
the scalar Laplacian applied to each vector component.

The terms in the definition for the vector Laplacian can be written
out as follows:

𝐀= ∇×𝐅
Φ= ∇⋅𝐅
𝐁= ∇×𝐀= ∇×(∇×𝐅)
𝗘=−∇Φ= −∇(∇⋅𝐅)

And because of vector identities, one can also write:

∇×𝗘= 0
∇⋅𝐁= 0

As can be seen from this, the Laplacian establishes a Helmholtz
decomposition of the vector field 𝐅 into an irrotational or curl free
component 𝗘 and a divergenceless component 𝐁.

In Maxwell's equations, the curl of the electric field 𝗘 is defined by:

∇×𝗘= -∂𝐁/∂t,

which is obvious not equal to zero and therefore Maxwell's equations
cannot be second order spatial derivatives of any vector function 𝐅
as defined by the Laplacian. On other words, herewith we have shown
that no function 𝐅 exists for which Maxwell's equations are the 2nd
order spatial derivative as defined by the vector Laplacian and
therefore the solutions of Maxwell’s equations do not satisfy
Laplace’s equation.

Furthermore, this violates the Helmholtz decomposition defined by the
Laplace operator and therewith it’s separation into an irrotational
and a divergenceless component. The end result of this is that while
the solutions of Laplace’s equations are all possible harmonic wave
functions, with Maxwell’s equations there is only resulting wave
function which defines a transverse wave, whereby the 𝗘and
𝐁components are always perpendicular with respect to one another.
This is also the reason why no separate wavefunctions can be derived
for the “near” and “far” fields.

Furthermore, in Maxwell’s equations, the two potential fields which
are used withHelmholtz’s theoremare the electrical potential Φand the
vector potential A, which are defined by the equations[i]:

B =∇×A,
E = −∇Φ−∂A/∂t

where B is the magnetic field and E is the electric field.

The Helmholtz theorem can also be described as follows. Let A be a
solenoidal vector field and Φ a scalar field on R3 which are
sufficiently smooth and which vanish faster than 1/r2at infinity. Then
there exists a vector field F such that:

∇⋅F=Φ and ∇×F=A,

and if additionally, the vector field F vanishes as r →∞, then F is unique[ii].

Now let us consider the units of measurement involved in these fields,
whereby the three vectoroperators used all have a unit of measurement
in per meter [/m].The magnetic field Bhas a unit of measurement in
Tesla[T], which is defined in SI units as [kg/s2-A].So, for the
magnetic vector potential Awe obtain a unit of [kg-m/s2-A]and for
dA/dt we obtain a unit of [kg-m/s3-A].The electric field Ehas a unit
of measurement involt per meter, which is defined in SI units
as[kg-m/s3-A], which matches that for dA/dt.So, for the electric
scalar potential Φwe obtain a unit of [kg-m2/s3-A].

Obviously,however, neither the units of measurement for E and B are
the same, nor are the units of measurements for Φ and A. This is in
contradiction with Helmholtz’s theorem, which states that a vector
field F exists that should have a unit of measurement equal to that of
 Φ and A times meters or that of E and B times meters squared.

Thus we have shown that Maxwell’s equations are incompatible with the
Laplace equation and are in contradiction with Helmholtz’s theorem as
well, which means that the potential fields defined by Maxwell are
inconsistent with the potential fields that are studied with potential
theory, the study of harmonic functions which are defined as the
functions that satisfy Laplace’s equation. And therefore Maxwell’s
equations should be revised.
-:-

Any suggestions on how I could improve this?

Best regards,

Arend.


P.S. Also see:

https://en.wikipedia.org/wiki/Second_derivative#The_Laplacian

"Main article: Laplace operator

Another common generalization of the second derivative is the
Laplacian. This is the differential operator  defined by

∇²f = ∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z².

The Laplacian of a function is equal to the divergence of the gradient
and the trace of the Hessian matrix."

https://en.wikipedia.org/wiki/Laplace_operator

"In mathematics, the Laplace operator or Laplacian is a differential
operator given by the divergence of the gradient of a function on
Euclidean space. "

https://en.wikipedia.org/wiki/Vector_Laplacian

"In mathematics and physics, the vector Laplace operator, denoted by
∇², named after Pierre-Simon Laplace, is a differential operator
defined over a vector field. The vector Laplacian is similar to the
scalar Laplacian. Whereas the scalar Laplacian applies to a scalar
field and returns a scalar quantity, the vector Laplacian applies to a
vector field, returning a vector quantity. When computed in
orthonormal Cartesian coordinates, the returned vector field is equal
to the vector field of the scalar Laplacian applied to each vector
component."



On Thu, Apr 23, 2020 at 5:18 PM Hans van Leunen <jleunen1941 at kpnmail.nl> wrote:
>
> Please read  https://www.researchgate.net/publication/339744488_Representing_basic_physical_fields_by_quaternionic_fields
> Hans
>
> Op 23 april 2020 om 17:13 schreef Maurice Daniel <5D at earthlink.net>:
>
> Arend Lammertink,
>
> If you are not already aware of the works of Professor Oleg D. Jefimenko, (1922 to 2009)  physicis and Professor Emeritus at West Virginia University; author of such works as: “Causality Electromagnetic Induction and Gravitation” then I suggest you read some of his books.  He has discussion of these topics, and like you, he has dispute with two of Maxwell’s equations.  He was also able to combine electromagnetism and gravity.
>
> I studied vector math at one time, but now I have forgotten most of it so I cannot follow your arguments.  It would be interesting to know if you have reached the same conclusions as Prof. Jefimenko.  If so, this would lend powerful support to your arguments.
>
> Let me know if this is helpful to you.
>
> - - - Maurice Daniel - - -
>
>
>
>
> Maurice Daniel
> 5D at earthlink.net
>
>
> On Apr 23, 2020, at 2:09 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:
>
> ∇²𝐅= ∇(∇·𝐅) - ∇×(∇×𝐅) = 0.
>
> This can be re-written as:
>
> -∇²𝐅= - ∇(∇·𝐅) + ∇×(∇×𝐅) = 0.
>
> Then, the terms in this equation can be written out as follows:
>
> 𝐀= ∇×𝐅
> Φ= ∇⋅𝐅
> 𝐁= ∇×𝐀= ∇×(∇×𝐅)
> 𝗘=−∇Φ= −∇(∇⋅𝐅)
>
> And because of vector identities, one can also write:
>
> ∇×𝗘= 0
> ∇⋅𝐁= 0
>
> So, any given vector field 𝐅 can be decomposed like this into a
> rotation free component 𝗘 and a divergence free component 𝐁.
>
> There is no argument this is mathematically consistent, nor that the
> solutions to the equation -∇²𝐅= 0 are the harmonic wave functions in
> 3D.
>
> Now compare this to Maxwell's:
>
> 𝗘= −∇Φ− ∂𝐀/∂t
>
> Take the rotation at both sides of the equation and we obtain the
> Maxwell-Faraday equation:
>
> ∇×𝗘= - ∂𝐁/∂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)—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 𝗘 to the time derivative of the divergence free (and
> therefore rotational) component 𝐁 by Maxwell results in 𝗘 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.
> <Revision of Maxwell equations DRAFT.pdf>_______________________________________________
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>
>
>
>
> Maurice Daniel
> 5D at earthlink.net
>
>
>
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