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

[Arend Lammertink]

On Fri, Apr 24, 2020 at 8:16 AM Ilja Schmelzer <ilja.schmelzer at gmail.com> wrote:
>
> 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.

Hmm. Should have been more precise when I said that. Wat I means was:

I agree that the very idea to prove that the Maxwell equations are
mathematically _internally_ inconsistent 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.

The point is that the Laplace operator IS what DEFINES the second
order spatial derivative within vector theory. There is no argument
the terms in this definition can be written out and labeled as
follows:

 ∇²𝐅= ∇(∇·𝐅) - ∇×(∇×𝐅)

The terms in this identity 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

And since in Maxwell's equations we have:

 ∇×𝗘= -∂𝐁/∂t,

which is not 0, and thereore there is also no argument that Maxwell's
equations are NOT  the second order spatial derivative of ANY vector
function 𝐅. The Laplacian is just the generalisation of the 1D second
order derivative ∂²/∂x² and indeed is elementary math.

I really don't see how one could possibly argue that a vector function
𝐅 exists, such that Maxwell's eqations form it's Laplacian and
therewith it's second order derivative.

The very fact that in Maxwell ∇×𝗘!=0 makes this undeniable, IMHO.


>
> > 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.

https://en.wikipedia.org/wiki/Lorentz_ether_theory
"Lorentz tried in 1899 and 1904 to expand his theory to all orders in
v/c by introducing the Lorentz transformation."

It's the very application of the Lorentz transformation to physics
that is the problem. And the reason it exists is because of, you
guessed it: Maxwell, or at the very least a misinterpretation thereof:

http://www.etherphysics.net/CKT4.pdf


>
> > 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.

https://en.wikipedia.org/wiki/Magnetic_vector_potential#Gauge_choices
"The above definition does not define the magnetic vector potential
uniquely because, by definition, we can arbitrarily add curl-free
components to the magnetic potential without changing the observed
magnetic field. Thus, there is a degree of freedom available when
choosing A. This condition is known as gauge invariance."

Would it be far fetched to suggest that the very reason the magnetic
vector potential has not been defined uniquely is because no vector
function 𝐅 exists, such that Maxwell's equations form it's Laplacian?


>
> > 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.

The gauge freedom within Maxwell is around the definition of the
vector potential, as quoted above:
"we can arbitrarily add curl-free components to the magnetic potential
without changing the observed magnetic field"

Could it be that because of the vector identity ∇⋅𝐁= 0 that "some
configurations of the field", namely the addition of curl-free
components to the field, by definition results in the zero vector?

>
> 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.

I respectfully disagree that there's no reason to care about the
origins of our theories and to check their validity.

Actually, I don't think it's an exxageration to state that this issue
defines the difference between a theory of everything and 150+ years
of trying to find additional equations in order to correct the obvious
violation of the elementary math defined by the Laplace operator.