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

[Arend Lammertink]

On Fri, Apr 24, 2020 at 2:59 PM Ilja Schmelzer <ilja.schmelzer at gmail.com> wrote:
>
> 2020-04-24 16:16 GMT+06:30, Arend Lammertink <lamare at gmail.com>:
> > On Fri, Apr 24, 2020 at 8:16 AM Ilja Schmelzer <ilja.schmelzer at gmail.com>
> > wrote:
>
> > 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.
>
> And I see no reason why one should make such an argument.

Ok, let's start with two possibilities of describing the fields within
an aether paradigm:

1) a basic fluid dynamics model for the aether (start with an ideal
Newtonian fluid) and on top of that a particle model;

2) Maxwell's model, with a/o a vector potential field that has not
been uniquely defined.


In fluid dynamics, we have both incompressible flow as well as
irrotational flow:

https://en.wikipedia.org/wiki/Flow_velocity
“Incompressible flow

If a fluid is incompressible the divergence of u is zero:

 ∇⋅u = 0.

That is, if u is a solenoidal vector field.



Irrotational flow

A flow is irrotational if the curl of u is zero:

 ∇× u = 0.

That is, if u is an irrotational vector field.

A flow in a simply-connected domain which is irrotational can be
described as a potential flow, through the use of a velocity potential
Φ, with u = ∇Φ. If the flow is both irrotational and incompressible,
the Laplacian of the velocity potential must be zero:
ΔΦ=0.”


Interestingly, in FD we do have a scalar velocity potential:

https://en.wikipedia.org/wiki/Velocity_potential
“A velocity potential is a scalar potential used in potential flow
theory. It was introduced by Joseph-Louis Lagrange in 1788. It is used
in continuum mechanics, when a continuum occupies a simply-connected
region and is irrotational. In such a case,

 ∇×u = 0

,where u denotes the flow velocity. As a result, u can be represented
as the gradient of a scalar function Φ”

And we have the Laplace equation for irrotational flow:

https://en.wikipedia.org/wiki/Laplace_equation_for_irrotational_flow
"Irrotational flow occurs when the cross gradient of the velocity or
shear is zero.

∇×v= 0

[…]

Now ∇×v=0 so,

[…]

These restrictions on the velocity must hold at every point. Consider
a function Φ which satisfies the condition

vdx + vdy = −dΦ

The minus sign is arbitrary it is a convention that causes the value
of Φ to decrease in the direction of the velocity. This proves the
existence of a function Φ such that its negative derivative with
respect to any direction is the velocity component in that direction.

v = −∇Φ

The assumption of a velocity potential is equivalent to the irrotational flow as

∇×(−∇Φ)=0

Whenever a velocity potential function exists then it must be an
irrotational flow so they are equivalent to each other.

[…]

In vector form it can be written as

 ∇²Φ = 0

It is called as the laplace equation. Any function Φ that satisfies
the laplace equation is a possible irrotational flow case. As there
are infinite number of solutions to the laplace equation each of which
satisfies certain flow boundaries the main problem is the selection of
the proper function for the particular flow case.“


But I have not been able to find anything involving a vector
potential. The closest I have been able to find are stream functions:

https://en.wikipedia.org/wiki/Stokes_stream_function
“In fluid dynamics, the Stokes stream function is used to describe the
streamlines and flow velocity in a three-dimensional incompressible
flow with axisymmetry. […] The velocity field associated with the
Stokes stream function is solenoidal—it has zero divergence.”

And there is vorticity:

https://en.wikipedia.org/wiki/Vorticity
“More precisely, the vorticity is a pseudovector field [ω], defined as
the curl (rotational) of the flow velocity [u] vector. The definition
can be expressed by the vector analysis formula:

[ω] ≡ ∇ ×[u]

where ∇ is the del operator. The vorticity of a two-dimensional flow
is always perpendicular to the plane of the flow, and therefore can be
considered a scalar field. As a consequence, the unit of the vorticity
is 1/s, i.e. Hz.”

But this is not a vector potential, it’s a derivative of the velocity field.


So, unfortunately, the use of potential theory within the area of
fluid dynamics has not been fully developed, the use of a vector
potential appears to be absent.


Now let’s head over here:

https://en.wikipedia.org/wiki/Aharonov–Bohm_effect#Significance
“Eventually, a description arose according to which charges, currents
and magnets acted as local sources of propagating force fields, which
then acted on other charges and currents locally through the Lorentz
force law. In this framework, because one of the observed properties
of the electric field was that it was irrotational, and one of the
observed properties of the magnetic field was that it was
divergenceless, it was possible to express an electrostatic field as
the gradient of a scalar potential (e.g. Coulomb's electrostatic
potential, which is mathematically analogous to the classical
gravitational potential) and a stationary magnetic field as the curl
of a vector potential (then a new concept – the idea of a scalar
potential was already well accepted by analogy with gravitational
potential). The language of potentials generalised seamlessly to the
fully dynamic case but, since all physical effects were describable in
terms of the fields which were the derivatives of the potentials,
potentials (unlike fields) were not uniquely determined by physical
effects: potentials were only defined up to an arbitrary additive
constant electrostatic potential and an irrotational stationary
magnetic vector potential.”


So, on the one hand we see that potential theory has not been fully
developed in fluid dynamics, and on the other hand we see that the
potentials are not uniquely defined within the electromagnetic domain,
but we also see the clear separation into a irrotational component [E]
and a divergenceless component [B], albeit in the static case.

So, in the static case, whereby -∂𝐁/∂t = 0, the argument I made no
longer holds and we can conclude that a vector field 𝐅 _does_ exist,
such that the following all fits tightly together:

 ∇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


Now if we go the the fluid dynamics domain for a moment, it is common
to work with the velocity field [v], both in the cases of
incompressible flow as well as irrotational flow, so the unit of
measurement for both 𝗘 and 𝐁 would be in [m/s]. And because all
three differential operators have a unit of measurement in per meter
[/m], we obtain a unit of measurement in square meter per second
[m^2/s] for the potential fields and cubic meters per second [m^3/s]
for 𝐅.

This would define a vector field denoting something like a volumetric
flow rate or volume velocity, a volumetric flux. And when you try to
integrate this flux over a closed volume and take the limit to zero,
you find that in the limit this flow rate goes to zero.

In other words: dimensional analysis reveals that in the case of fluid
dynamics, we have:

∇²𝐅= ∇(∇·𝐅) - ∇×(∇×𝐅) = 0,

by definition. This is rather remarkable, since this equation (or
whatever you want to call it) is what’s known as the/a Laplace
equation.


Now to answer the question: it seems to me that with a little more
puzzling, we can work out a complete theory that fits like a glove,
were it not for Faraday's law. And as I argued in my reply to Daniel,
IMHO there are ample reasons to introduce Faraday's law somewhere else
in the model.

>
> First of all, the Laplacian of a function is another function,
> and not an equation.  But, whatever this means, why would somebody
> claim that, say, solutions of the Maxwell equations are second derivatives
> of whatever?

May be I use these words in a confusing way, but the point is the following:

In vector theory, the Laplacian is essentially a generalization of the
second order derivative ∂²/∂x² into three dimensions.

And since the dimensions associated with the three operators used are
in [/m], the Laplacian can be considered to be an operator that has a
dimension (unit of measurement) in per meter squared [/m^2].

So, it seems pretty clear that the Laplacian defines a 2nd order
spatial derivative with a dimension in [/m^2].

And if this operator IS a (spatial) derivative, there should be
something where the resulting field/function/whatever IS the
derivative OF and that something should exist.

>
> > 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.
>
> No, it is not a problem at all.
>

There is no argument that the application of the Lorentz transform is
what changes "flat spacetime" into "curved spacetime":

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

"Minkowski's geometric interpretation of relativity was to prove vital
to Einstein's development of his 1915 general theory of relativity,
wherein he showed how mass and energy curve flat spacetime into a
pseudo-Riemannian manifold."

>
> > 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?
>
> First you would have to suggest something which transform a vector function
> into some equations (instead of, say, another vector function like the
> Laplacian)
> and is nonetheless named "Laplacian".

The point is:

1) The Laplacian defines some kind of derivative;

2) Locically, therefore, something should exist where the Laplacian
defines the derivative of.

Just think of integration: finding an Integral is the reverse of
finding a Derivative.

Or to put it the other way around: if a derivative exists, it's
reverse should also exist.

That's the idea. I know it doesn't always count, this is just to get
the idea across.


>
> > 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?
>
> Again, you have to explain what you mean, given that addition is
> an operation, but a configuration is not.

Elsewhere, I made this argument:

Gauge theory is built on the principle that you can add curl-free
components to the vector potential field [A] and divergence free
components to the scalar potential field Φ. Vector theory learns that
doing so results in the null vector for the resulting force.
Therefore, we can conclude that gauge theory yields no force and
therefore no physical effect at all. It should thus be rejected and
therefore QFT should be revised.

It's the same vector identities that enable us to write:

 ∇×𝗘= 0
 ∇⋅𝐁= 0

that lead to the conclusion that neither the addition of a curl-free
component to the vector potential field [A] nor the addition of a
divergence-free component to the scalar potential field Φ result in
anything other than zero in the respective force fields 𝗘 and 𝐁.

These identities are:

https://en.wikipedia.org/wiki/Vector_calculus_identities#Second_derivative_identities

The divergence of the curl of any vector field A is always zero:
∇⋅(∇×A)=0

and

The curl of the gradient of any continuously twice-differentiable
scalar field  is always the zero vector:
∇×(∇Φ)=0

>
> > I respectfully disagree that there's no reason to care about the
> > origins of our theories and to check their validity.
>
> Fine, check whatever you like to check,
> But what follows if it appears that the origins are nonsense?

What follows if it appears that what came after the origins is even
bigger nonsense?

>
> The theory may be nonetheless meaningful and make correct
> predictions.  Kepler believed in astrology, and his origins were
> astrological nonsense.  But his laws of motion for the planets were
> nonetheless very good approximations.
>
> > 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.
>
> I see up to now only confusion. Sorry.

Hope this helps in clearing things up.

Best regards,

Arend.


>
> Greetings Ilja
>
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