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

[Hans van Leunen]

See
http://www.e-physics.eu/__QuaternionicDifferential.rtf
it contains: 

Quaternionic differential calculus applies the quaternionic nabla operator ∇. This calculus uses proper time τ, where Maxwell equations apply coordinate time t. 
We use boldface to indicate vectors.
∇≡∇ᵣ+∇
∇≡{∂/∂x, ∂/∂y, ∂/∂z}
∇ᵣ≡∂/∂τ
In the quaternionic differential calculus differentiation is a multiplier operation.
Compare:
c=cᵣ+c=a b≡(aᵣ+a)(bᵣ+b)=aᵣbᵣ−〈a,b〉+abᵣ+aᵣb±a×b
and
ϕ=ϕᵣ+ϕ=∇ψ≡(∇ᵣ+∇)(ψᵣ+ψ)=∇ᵣψᵣ−〈∇,ψ〉+∇ψᵣ+∇ᵣψ±∇×ψ
∇ψ represents the change of field ψ. This change covers five terms. Part of these terms can compensate for other terms. For example, ∇ᵣ ψ can compensate ∇ψᵣ.
ϕᵣ=∇ᵣψᵣ−〈∇, ψ〉
Maxwell does not use the real part ϕᵣ of ϕ
ϕ=∇ψᵣ+∇ᵣ ψ ±∇×ψ
Double differentiation leads to the quaternionic second order differential equation:
ζ=∇*ϕ=(∇ᵣ−∇) (∇ᵣ+∇) (ψᵣ+ψ)= {∇ᵣ∇ᵣ+〈∇, ∇〉} (ψᵣ+ ψ)=ρᵣ+J
ζ={∇ᵣ∇ᵣ+〈∇, ∇〉} ψ
The equation can be split into two first-order partial differential equations ϕ=∇ψ and ζ=∇*ϕ
This equation offers no waves as its solutions. The quaternionic equivalent of the wave equation is:
φ= {∇ᵣ∇ᵣ−〈∇, ∇〉} ψ
The homogeneous version of this equation offers waves as its solution:
∇ᵣ∇ᵣ ψ=〈∇, ∇〉ψ =−ω² ψ
Corresponding Maxwell-like equations are:
E=−∇ψᵣ−∇ᵣψ
B=∇×ψ
ρᵣ=〈∇,E〉
J=∇×B−∇ᵣE
∇ᵣB=−∇×E
The corresponding second-order differential equations are:
{∇ᵣ∇ᵣ −〈∇,∇〉}ψ=J
{∇ᵣ∇ᵣ −〈∇, ∇〉}ψᵣ=ρᵣ

In regular physics
Momentum p=h∇ 
Kinetic energy T = (p²/2m) =−(h²/2m)〈∇, ∇〉
Total energy:Hamiltonian H=T+V=(h²/2m)〈{∇ᵣ∇ᵣ−〈∇, ∇〉}
Lagrangian L=T−V=− (h²/2m){∇ᵣ∇ᵣ+〈∇, ∇〉}

Solutions
The two second-order partial differential equations describe the behavior of dark objects.
φ=(∂²/∂τ²± 〈∇, ∇〉)ψ
A third equation skips the first term
φ= 〈∇, ∇〉ψ
This is the Poisson equation.
In fact, these equations are quaternionic differential equations. Thus, φ and ψ are quaternionic functions that own a scalar real part and an imaginary vector part. The solutions are quaternionic functions.
The equation using the − sign is the quaternionic equivalent of the wave equation. The equation using the + sign splits into two quaternionic first order partial differential equations. This second equation does not offer waves as solutions.
The homogeneous wave equation can be written as: 
∂²ψ/∂τ²= 〈∇,∇〉ψ =-k² ψ
This delivers the quantized wave solutions of the Helmholtz equation
The dark objects behave as shock fronts and operate only as odd-dimensional field excitations. During travel, all shock fronts keep the shape of the front.
All solutions of a homogeneous second-order partial differential equation superpose in new solutions of that equation
One-dimensional shock fronts
The one-dimensional shock fronts also keep their amplitude. Consequently, the one-dimensional shock fronts can travel huge distances without losing their properties. Combined equidistantly in strings, they represent the functionality of photons. This means that the one-dimensional shock fronts are the tiniest possible packages of pure energy.
Depending on the PDE, the solutions can be described by different equations. The solution for the wave equation is
g(q ,τ)=f(c τ ± |q−qₒ|)
This solution cannot represent polarization.
The solution for the other equation is
g(q ,τ)=f(c τ ± |q−qₒ| i)
The vector i can indicate the polarization of the shock front.
This solution represents a dark energy object.
A photon is a string of equidistant energy packages that obeys the Einstein-Planck relation
E = h ν.
Since photons possess polarization, they use the second solution for their energy packages. Thus, the constituents of photons are not solutions of the wave equation.
Green’s function
One of the solutions of the Poisson equation is the Green’s function
g(𝐪)=1/|𝐪−𝐪ₒ|
∇g(𝐪)=( 𝐪−𝐪ₒ)/|𝐪−𝐪ₒ|³
〈∇, ∇〉 g(𝐪)= 〈∇, ∇g(𝐪)〉=4πδ( 𝐪−𝐪ₒ)
Thus, the Green’s function is a static pulse response under purely isotropic conditions.
Three-dimensional shock front
The three-dimensional shock fronts require an isotropic trigger. These field excitations integrate over time into the Green's function of the field. That function has some volume, and the pulse response injects this volume into the field. Subsequently, the front spreads the volume over the field. The corresponding solution of the wave equation is
g(r ,τ)=f(c τ ± r)/r
The parameter r is the radius of the spherical front. The formula can also be written as
g(q ,τ)=f(c τ ± |q−qₒ|)/|q−qₒ|
The solution for the other PED is
g(q ,τ)=f(c τ ± |q−qₒ| i)/|q−qₒ|
In this solution, the vector i acts as a spin vector. It is a normed imaginary quaternion.
This solution represents a dark matter object.
Thus, the initial deformation quickly fades away. but the expansion of the field stays. Having the capability to deform the carrier field corresponds to owning a corresponding amount of mass. This means that temporarily, the spherical pulse response owns some mass. This mass vanishes, but the expansion stays.
A huge coherent, recurrently regenerated swarm of spherical pulse responses can generate a significant and persistent deformation that moves with the swarm. This happens in the footprint of elementary particles. The spherical pulses are generated by the hop landing locations of the particle. The hopping path forms a hop landing location swarm that is described by a location density distribution. This distribution equals the square of the modulus of the wavefunction of the particle.
Pulse responses are a solution of {∇ᵣ∇ᵣ±〈∇, ∇〉} ψ(𝐪, τ)=4πδ( 𝐪−𝐪ₒ)θ(τ−τₒ)


Or read https://www.researchgate.net/project/The-Hilbert-Book-Model-Project/update/5e81bdc1498d5000016e0c71

Greathings, Hans van Leunen

> Op 24 april 2020 om 8:15 schreef Ilja Schmelzer <ilja.schmelzer at gmail.com>:
> 
> 
> 2020-04-24 12:16 GMT+06:30, Arend Lammertink <lamare at gmail.com>:
> > On Fri, Apr 24, 2020 at 6:42 AM Ilja Schmelzer <ilja.schmelzer at gmail.com>
> >> The very idea to prove mathematically that the Maxwell equations are
> >> incorrect makes no sense.
> >>
> >
> > I agree that the very idea to prove that the Maxwell equations are
> > _mathematically_ incorrect makes no sense, I'll give you that.
> >
> > The point is that Maxwell's equations are totally out of whack with
> > the vector Laplace equation. How much sense does that make?
> 
> Also not much.  These are quite elementary mathematics, so either
> the result is completely irrelevant or you made some elementary error.
> 
> > Especially given that Maxwell's equations eventually lead to both
> > relativity as well as gauge theory and QFT, two mutually exclusive
> > theories that cannot both be correct.
> 
> If you have in mind special relativity, the only conflict with QFT appears
> if you insist on the Minkowski spacetime interpretation. With the Lorentz
> ether interpretation you cannot prove the Bell inequalities (quantum causal
> influences could go FTL in the hidden preferred frame) so that no
> conflict occurs.
> 
> > Hasn't the "gauge freedom", which forms the basis for gauge theory,
> > crept into the model exactly because Maxwell's equations are totally
> > out of whack with the vector Laplace equation?
> 
> Whatever language you use to describe this ("crept into", "out of whack")
> it does not change the well-known and established facts about the
> Maxwell equations, in particular their gauge invariance.
> 
> > How much sense does it make to have a model which has been developed
> > within the aetheric paradigm that has "gauge freedom"?
> 
> Gauge freedom simply means that some configurations of the field
> cannot be distinguished by measurements.  This is completely unproblematic
> except for empiricism/positivism, where unobservable things cannot exist.
> 
> But there is no reason to care about these outdated philosophies of science.
> In Popper's critical rationalism restricted human abilities of
> observation are not
> a problem at all.
> 
> In my approach to ether theory in modern physics gauge fields play a central
> role too.  See https://ilja-schmelzer.de/matter
> 
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