[Physics] Do longitudinal FTL "Tesla" waves exist and, if yes, how should they be modelled?

[Ilja Schmelzer]

2020-05-06 4:35 GMT+06:30, Arend Lammertink <lamare at gmail.com>:
> On Tue, May 5, 2020 at 12:23 AM Ilja Schmelzer <ilja.schmelzer at gmail.com>
>> No, the mainstream hopes a lot to unify them, but has failed up to now.
>
> The alternative view is that there is only one fundamental interaction
> of Nature, namely the electromagnetic domain. From that perspective,
> it is hopeless to try and fix things before fixing the electromagnetic
> domain model aka Maxwell's equations.

Given the SM, it seems quite strange to think that the EM field is
somehow fundamental.

>> I think this is hopeless.
>
> From my perspective, it is inevitable.
>
> Once one realizes how close Maxwell's equations actually are to a
> fluid dynamics model describing motion in a fluid-like medium called
> aether and one compares Maxwell's model to LaPlace / Helmholtz math,
> it is obvious that the term dB/dt is where Maxwell's equations
> differentiate with the fundamental theorem of vector calculus.
>
> I don't think there can be any disagreement about this fact.

There obviously is.  As explained, you cannot get rid of the dB/dt
term without destroying the whole theory, and it follows simply that
there is no closeness.

> And I also don't think there can be any disagreement about what it is
> that is being described by the equation curl E = -dB/dt: Faraday's
> law.
>
> So, the disagreement comes down to the following questions:
>
> Is Faraday's law a relation that holds on a fundamental level?

No, this is not the question.  The first question is if curl E = 0 is
viable at all given Faraday's experiment.

In my ether theory, it is not a law on the fundamental level (where we
have a discrete version of all the equations). Before caring about the
fundamental level, one has to accept that there should be some limit
where Faraday's law holds. This rules out curl E = 0.

> Is it absolutely correct that in the case of varying fields (waves)
> these two fields *must* always be perpendicular to one another, no
> matter what?

They must not.

> The experimental verification of the existence of a FTL wave within
> the electromagnetic domain would prove that Faraday's law is not a law
> that applies at the fundamental level. It would prove that equating
> curl E to -dB/dt at a fundamental level in the model is incorrect. It
> would prove that the elemental math as defined by LaPlace / Helmholtz
> also applies within the electromagnetic domain.

First, math always applies everywhere.  Then, what you apply here is
not math, but a particular idea about an ether theory which is not
viable because curl E = 0 is not viable.

>> Sorry, once some theoretician has found a way which has some chance of
>> success, he no longer needs opening his mind to whatever alternatives.
>
> That would almost be a contradiction in terminus. Unless he is
> absolutely certain of success, he cannot rule out that some
> alternative actually offers a better chance of success and therefore
> he must keep an open mind for such a possibility.

No, this may be a purely pragmatic choice. Yes, even now I cannot
exclude that other approaches may be better than what I have found.
Maybe it is nonetheless string theory?  So what, that would be bad
luck for me, that's all.

>> Such "opening mind" exercises may be useful for those who want to work
>> in fundamental physics but have no idea how to start.

> What I've been trying to tell you all this time is this: start with a
> simple fluid-dynamics based aether model as your fundamental model and
> compare that with Maxwell's.
> What does stand out?
> What does that tell you?

I see an immediate failure of your simple ether model. I see a viable
more complex model which gives almost the whole SM and GR, inclusive
the Maxwell theory.

> Is it really far fetched to suggest that the way Maxwell deviated from
> fundamental, elemental math was, in actual fact, a gigantic blunder?

Yes. To suggest that the Maxwell equation deviated from math is simply
complete nonsense, I have tried to show you a variant which makes at
least sense, namely that the Maxwell equations are in conflict with
your extremely simple ether model.

>> I have survived nicely without own data. I had, with some luck, a
>> guiding idea which put me on the way to develop an ether theory. It
>> had already from the start the necessary equations
>
> What I'm offering is exactly such a guiding idea, namely that this
> equation actually means something:
>
> ∇²𝐅= ∇(∇·𝐅) - ∇×(∇×𝐅) = 0

Feel free to speculate about the meaning of this. I think the very
idea is nonsensical.

> Bear in mind that the development of the SM was guided by the idea
> that there was "gauge freedom" in Maxwell's equations.

This was not an idea, this was and is a simple mathematical fact about
these equations.

> What if Maxwell indeed made a blunder and this whole "gauge freedom"
> idea was in fact just an illusion?

The Maxwell equations, as equations for E and B, predict a lot of
things about observables, and these predictions have been tested a lot
of times. This agreement between the theory and observation is
certainly not just an illusion, it is a very strong hard fact.

This fact is so hard that you are essentially forced, if you modify
the Maxwell equations, to show that in the region where it has been
well-tested they hold approximately.

>> Yes, that's what has to be expected. My SM model is also not exactly
>> the SM. It is, roughly, as close to the SM as possible (and necessary
>> to be viable).
>
> Yes, in order to build a viable model, one has to retain the
> predictions made by competing models in one way or the other. So, if
> one wants a chance of success, one is pretty much limited to
> refactoring / rearranging the existing model, but one can't just throw
> it away and start from scratch.

Fine, let's remember this.

>> No. There can be many many failures. And looking at how some guy
>> performes some experiment would not be the appropriate way of error
>> search.
>
> That is true, but the whole idea behind physics is that mother Nature
> does not fail to react in exactly the same way
> when one performs exactly the same experiment.
>
> In that sense, Wheatsone's experiment is once again very interesting.

Feel free to be interested and to repeat it.  That's not my problem,
and I cannot support you here. But what I can see is that your curl E
= 0 idea is completely off because it destroys the Maxwell equations
completely, with no chance to recover it in any limit.

>> Who knows?  But I doubt that such a classical mechanism can be of any
>> use, given that QT predicts all these things nicely.
>
> Doubt is good. It means one can't rule it out, either, and therefore
> the mind is still open for the possibility.

That's a triviality, one can never rule out that some other theory is
right and the own theory fails.  Such is life. This does not mean that
there is much of an open mind - one will not spend much own time in
hopeless things.

>> You have not yet a theory (with evolution equations and so on) which
>> gives these waves.
>
> I agree I don't have a quantifyable theory, but I do have the
> fundamental idea that essentially defines the fundamental foundation
> for a quantifyable theory in one equation:
>
> ∇²𝐅= ∇(∇·𝐅) - ∇×(∇×𝐅) = 0

This is simply nothing.

> All that in essence stands in
> between this approach and Maxwell's is Faraday's law.

And that's the end of the story for this approach. It is dead from the
start, falsified by Faraday's experiment.

> So, come to an agreement that Faraday's law has indeed been introduced
> into the model at a place where it does not belong and we are looking
> at a revolutionary different foundation for theoretical physics,
> namely the idea that the aether actually behaves like a fluid and
> should therefore be described as such.

No. The Maxwell equations for E and B are a nice place, because both E
and B can be and have been measured.

>> According to EM theory (Maxwell) it is simply impossible as to create
>> them as to measure them.
>
> Yep.
>
>> You have not yet another theory.  Except the
>> one where dB/dt is simply removed, which is as dead as possible. And
>> that "theory" will not give any waves at all.

> In my view fluid dynamics IS a theory.

It is a theory about fluids, not about the EM field.
Your ether theory where dB/dt is simply removed is also a theory, but
one which is empirically falsified by Faraday's experiment.

> Just fill in the right
> parameters like density and elasticity and there you have your aether
> model. That's it, nothing more to it than that.

Except that you have to make the right guesses, else the theory simply
fails, and that's it.  Moreover, the idea that the ether is fluid may
be completely wrong, it may be a solid or a plasma or whatever else.
In my theory, it is quite solid.

> So, what I'm actually saying is that you have all of the phenomena
> known in fluid dynamics, including waves, when you describe the aether
> as an ideal, Newtonian fluid. So, without working things out, one can
> come to conclusions like that a longitudinal wave will propagate a lot
> faster than a "transverse" wave.

If you have a liquid, you simply have no transverse waves.

> So, I'm not saying "just remove the dB/dt term and that's it", I'm
> saying: return to a FD model wherein you describe the aether as an
> ideal, Newtonian fluid and that the term dB/dt is the main obstacle in
> our way.

So your curl E = 0 ether theory is dead?  Fine. But, it seems, it is
yet alive in your mind:

> In other words: all that stands in between a fluid-dynamic model for
> the aether and classic electrodynamics is the way Maxwell described
> Faraday's law by the introduction of the dB/dt term at a place where
> it does not belong.

But it is at a place where you can explicitly make predictions about
observables, and then measure these observables, as Faraday has done.

>> > I think he would also like Occam's razor.
>>
>> Of course. But that does not mean that he would reject established
>> equations which make a lot of well-tested predictions.
>
> Certainly. But I doubt he would object to re-arranging such well
> established equations such that they fit with a model derived from a
> single fundamental hypothesis:
>
> The aether behaves like a fluid and should therefore be described as such.

Yes, that would be fine.  But you have to rearrange them in such a way
that the original testable predictions remain unchanged.

> The question is why you believe it is not tenable and no waves would
> be possible,

By looking at the equations proposed. But forget about the waves which
are impossible with your equations, just focus on Faraday's experiment
which falsifies curl E = 0.

>> First of all, you must recognize that the remaining theory is false
>> and can easily be falsified.
>
> Would be interested in such a falsification, I don't see it.

The electric field predicted for Faraday's experiment would be
curl-free, and, therefore, would be unable to create a current in a
closed loop.

>> The default answer is "look at wikipedia". For the information how to
>> measure it this should be sufficient. The result will be quite
>> obvious. Namely \nabla \times \mathbf {E} = 0 is dead.
>
> The correct answer is: virtually noboby has a freakin' idea!

So what? It does not matter, given that we have devices which measure E and B.

> Remember what you wrote earlier?
>
> "People have started with abstract fields in thermodynamics,
> and then, based on the atomic theory, have learned how these
> observable phenomenological fields depend on the properties of the
> atomic models.  This research program was successful in thermodynamics
> as well as in condensed matter theory."
>
> Maxwell started the same way, by introducing an abstract quantity
> called "electric charge".
>
> Only, in this case it has never been satisfactory explained what that
> actually is,

But this is not necessary to test particular equations. For testing
how the temperature changes we need a thermometer, not a theory about
the fundamental nature of temperature.

> In a nutshell: EITHER the particles cause the fields OR the fields
> cause the particles, but NOT both at the same time!

In a nutshell a phenomenological theory will not tell you what is
cause and what is effect.  It describes the fields we can measure, and
is based on the definition how they can be measured (with certain
measurement devices). A theory which introduces some causal
explanation would have to care about such things, but the Maxwell
equations, as equations for E and B, are a phenomenological theory
about those two fields E and B which can be easily measured, and does
not contain speculations about causal relations.

That popular explanations on wiki level contain causal ways to
describe some aspects of these equations is quite irrelevant.

> It is interesting and necessary in order to put the \nabla \times
> \mathbf {E} = 0 if dB/dt is nonzero into proper perspective.
> ...
> So, it is very important to take this point home: For an ideal coil,
> having zero resistance and zero parasitic capacitance, there is zero
> voltage and a zero electric field!

But zero resistance is a quite uninteresting limiting case. And we
don't have to care about this strange limiting case with no electric
field, given that we would like to measure the electric field.  One
way to measure an electric field is, clearly, to use a wire with some
resistance and measure the resulting current. Your ideal wire simply
distorts the E field, so it is inappropriate for measuring it.

>> No. You already have a problem, namely an experiment where dB/dt is
>> nonzero and, as a consequence of the Maxwell equations, \nabla \times
>> \mathbf {E} =/= 0.  And where all you have to do is to measure the
>> electric field in this situation to see that really \nabla \times
>> \mathbf {E} =/= 0. This is the decisive experiment between Maxwell's
>> theory and your "theory".
>
> What is decisive is the consideration of what it is that causes curl E
> =/= 0 in a practical experiment.
...
> So, let's once again draw in the analogy of what we're actually
> looking at with Faraday's experiment: a magnetic vortex, which is
> rather interesting, since there's a very interesting detail around the
> theoretical irrotational vortex I hadn't noticed before:

No, I couldn't care less about your vortexes, whatever they are. I
care about the electric field. Once an ideal coil simply distorts the
E field too much, I would suggest not to introduce them.

> So, yes, for this particular experiment that relationship is: curl E =
> -dB/dt and it holds up to rather high frequencies for practical coils,
> BUT that in no way implies that this is a fundamental relationship
> that ALWAYS holds and THAT's the whole point!

No, that's not the point.  It is quite sufficient to have a _single_
experiment where curl E = -dB/dt =/= 0 to show that the theory that
curl E = 0 is dead. And this is the point I care about here and now.

> This once again begs the question: what IS charge?
>
> Why is it on the one hand a property of certain "charged" particles
> yet at the same time a fundamental quantity that causes the fields,
> which makes that it becomes impossible to consider the possibility
> that "particles" are actually caused by the fields as well?

Before caring about such speculative questions, one has to get the
equations straight.  And to reject nonsense like curl E = 0 as a
general equation once we have found situations where curl E = -dB/dt
=/= 0.

>> Ok, but if there is a theory consistent (for those low frequencies)
>> with the experiments, and you don't question the experiments, you have
>> to be able to recover, in your modified theory, the successful
>> predictions of the old theory you have questioned.
>>
>> But you fail. For Faraday's experiment, your \nabla \times \mathbf {E}
>> = 0 equation predicts no current, but Faraday has observe one.
>
> It's actually the other way around: the relationship describing how an
> ideal coil interacts with a magnetic flux is what predicts a current,
> but no voltage and no electric field.

We don't care about ideal coils, we care about Faraday's experiment.

> The electric field is being observed, yes, but that's because in
> practice one cannot have an ideal coil and neither can one have an
> incompressible medium and therefore a pressure gradient will be
> observed in practice, which is what we call the electric field.

Whatever, once we have found situations where curl E = -dB/dt =/= 0
the theory curl E = 0 is dead.

What's the problem with acknowledging this?

>> Don't distract. If it fails to recover the result for the Faraday
>> experiment, it is dead, and nobody cares about what it thinks about
>> those hypothetical anomalies.
>
> The result for Faraday's experiment can be easily explained by
> starting out at the equation for an ideal coil and considering why
> this in practice leads to the presence of an electric field as well.

But I'm not interested in a theory about what happens inside ideal
coils, that's the theory of superconductivity, but in a theory about
the EM field. The E field is simply trivial inside, the magnetic field
will be expelled by the Meissner effect,
http://www.supraconductivite.fr/en/index.php?p=supra-levitation-meissner
so that the result is a trivial theory inside, and this thing cannot
test dB/dt =/= 0.

But, ok, no problem, I admit that your theory curl E = 0 is viable
inside a superconductor where we have E = B = 0, and, therefore, also
dB/dt = 0 so that the Maxwell equations hold too.

Let's now stop to consider superconductivity and handle a usual
vacuum, using the forces acting on charged kork balls to measure E and
using wires only to create a variable B.  Or with wires which have a
resistance so that the resulting currect can be used to measure the E
field.

> What's problematic is enforcing this result at the fundamental level
> in your model such that it HAS to apply exactly like this for all
> possible experiments which involve either a changing electric or a
> changing magnetic field.

Yes. The starting point would be to accept the Maxwell equations as
they are, as phenomenological equations for E and B. Which, if
modified, have to be modified in such a weak way that they can be
easily recovered in some limit. And, as a consequence, to throw away
ideas about ether theories which are unable to reach this, because the
E field would have to follow the equation curl E = 0.

> In practice, one gets away with this for a very long time, but it is
> clear that by doing so one makes it next to impossible to analyse the
> single wire transmission line, as illustrated by Elmore. One cannot
> calculate an impedance for this case, even though it has been found to
> equal the impedance of free space (377 Ohms) and therefore Maxwell's
> equations clearly break down here and need to be worked around.

That's your hypothesis that it breaks down. I doubt, and would prefer
to wait how the mainstream describes this. Whatever, this is not the
point I care about here, this can be left to future research.  The
thing I care about is that your curl E = 0 theory is dead, given
Faraday's experiment, and that you up to now refuse to acknowledge
this quite obvious and trivial thing.

>> Up to now, you have not found a viable way to rearrange something.
>> \nabla \times \mathbf {E} = 0 is in conflict with Faraday's
>> experiment.
>
> Faraday's experiment can be fully explained using physics based on the
> assumption of the existence of a fluid-like aether and therefore there
> is no actual conflict.

No. You have not given such a full explanation.

> In actual fact, it is the introduction of the term dB/dt into a
> fluid-dynamic model that is conflicting with the elemental math as
> defined by LaPlace / Helmholtz. It is really a bad idea to write
> equations that are in conflict with a fundamental mathematical
> theorem.

Again you fall back into complete nonsense. Nobody introduces
something into your fluid-dynamic model, it simply fails, because in
reality we have Faraday's experiment where dB/dt  =/= 0.  If your
fluid-dynamic model does not survive the introduction of the term
dB/dt, that fluid-dynamic model is simply dead. Big deal. Learn to
live with this, I have tried hundreds of ideas and had to throw them
away because they did not work.

>> Whatever, we have a force acting on small charged kork balls, not?
>> And we can measure this force, by putting such kork balls at some
>> interesting places, not?  This force field is known as E, and it is
>> not a good idea to redefine it.
>
> Actually, the units of measurement within the electromagnetic domain
> are undefined, except in relation to one another.
>
> The SI unit for electric field strength is volt per meter [V/m]
> The Volt is defined as [J/C] or [kg m^2 / A s^3], so the unit of
> measurement for E equals [kg m / A s^3].
>
> The Coulomb is defined as [A s], while the Ampere is defined as [C/s],
> so actually these units of measurement are only defined in relation to
> one another phenomenologically and therefore it might be an excellent
> idea to actually define what charge is and what current is and I think
> I finally figured out the correct way to do it.

It does not matter at all to write down the units. What the SI defines
is how these things are measured.  So learn how the SI works, what it
defines and how, namely be defining particular measurement procedures
for each unit.

The SI definitions make a lot of sense, because they are based on the
most accurate measurement procedures for each unit. Once experimental
science makes an advance, creating a device which measures some unit
more accurate then the old standard, they change the definition and
base the new definition on the new device. For this purpose, they
measure the old standard of what is 1 unit many times with the new
device, and use the result to define the same 1 unit now with the new
measurement device.  For the usual applications nothing changes,
because the extreme accuracy is not necessary for them anyway, and
they don't have to bother. 1 A remains 1 A, the old Amperemeter works
as before.

Your proposal seems unaware of those basic ideas of the SI system, so
I will simply ignore it.

>> > Bottomline is: when you revise Maxwell's equations, everything changes
>> > within theoretical physics.

>> No. All the experiments remain the same, with the same results. You
>> may somehow reinterpret something, but not that much. Revising the
>> Maxwell equations is certainly not a good idea, they can be easily
>> tested in many details.

> I did say *theoretical* physics. In the end, everything is based on
> Maxwell, one way or the other. So, if you change that, a lot of people
> are going to have a lot of work.

Theoretical physics has to care about predicting experimental results,
and interpreting experimental results too.

And as long as you care about things which can be directly measured,
like E and B, to change the equations is possible only if you recover
the well-established well-tested equations in a limit.  In this case,
not that much changes: Whenever that limit is sufficient, given the
accuracy requirements, you can use the old equations.

> But again the point: Yes, they can be easily tested in many details
> and the observations match, PROVIDED you stay within the fundamental
> concept of having CIRCULAR currents forming a closed loop, as ALL of
> our electronics do.

No. You can test the EM equations also in situations where you have no
wires at all. Use charged kork balls and measure the forces acting on
them.

>> In the low frequency domain, say, in Faraday's experiment, waves play
>> no role at all.
>
> Well, the electric field emitted by an electron in the shape of a
> longitudinal wave has a frequency of about 175 GHz.

Means it is irrelevant for Faraday's experiment.  No reason to care.
Your theory fails to predict the outcome of Faraday's experiment.

>> And that closed loop circuit is not necessary. Use
>> kork balls to identify the direction of the force. To falsify  \nabla
>> \times \mathbf {E} = 0 they will be good enough:
>>
>> no force     force ->          force <-
>> |                \                  /
>> |                 \                /
>> o                  o              o
>>
>> Then, simply let the kork follow the direction of the force. If you
>> finally arrive at the starting point, you have falsified  \nabla
>> \times \mathbf {E} = 0. (Again, I don't worry about the details, you
>> would need some dB/dt ~ const some time, this is not what I care
>> about, these are theoretical thought experiments, and you have to care
>> about the practical details.)
>
> You have falsified nothing. All you have shown is that in practice the
> medium is not incompressible and that as a result of the rotating
> aether in the vortex also a pressure gradient arises which is needed
> to balance the centripedal force within the vortex.
>
> The point is: this is the result of the physics of a vortex and
> therefore Faraday's law does not belong in the eqations which describe
> the dynamics of the medium itself.

I have falsified the theory with curl E = 0.  Your talk about some
compressible or incompressible medium as well as about vortices is
completely irrelevant.  I'm talking about the E field and the result
of measuring it with charged kork balls.

>> Sorry, no. Don't look back to Wheatstone, look first back to Faraday.
>> Once you don't like it with measuring the current, ok, do it with kork
>> balls. This measures E more directly, by measuring the force acting on
>> those kork balls.
>
> No need, it can be easily explained with the physics of the vortex in
> combination with Ampere's original circuit law:
>
> J = curl B.

No. We have no circuit here, we have charged kork balls and an
electric force acting on them.

> If the magnetic field is not changing, you have an *irrotational*
> vortex aka a "static" magnetic field and that is why curl B = 0 and
> therefore why there is no current.
>
> As long as the magnetic field is changing, the vortex is no longer
> irrotational and thus you get a current.

Nobody cares about your vortices, we care about the E field and
measure it with charged kork balls.

> And because the medium is not incompressible, there is also an
> electric field. But that is the result of vortex physics and not a
> fundamental relationship between the E and B fields that should be
> part of the medium model.

These ideas about vortex physics or your medium are irrelevant.  We
have an experiment and measure the electric field, by measuring the
force acting on charged kork balls (quite inaccurate, but we need only
the direction of the force to see that curl E =/= 0).

And we do not care about fundamental speculations but about measurable
fields E and B in that variant of Faraday's experiment.

>> No. There is no such animal as such "proper relations", they may be
>> proper only in the context of a particular theory. Here, your ether
>> theory. But your theory is in no way relevant for the Maxwell
>> equations.  They are well-defined, are about observable fields E and
>> B, and are well-tested.
>
> I still think fundamental mathematical theorems matter, especially
> when they give you a hint about how to define things like "what is a
> current". I actually started out thinking from the unit of measurement
> of the [B] field when I wanted to understand what current is and found
> that way that current should have a unit of measurement in [Hz].
> That's what I needed to come to the realization that a real current is
> actually a movement of charge carriers and not of charge per se.

Whatever it is, E and B are well-defined and can be measured.

About mathematical theorems you have to care if you invent an ether
theory.  If they tell you that in your ether theory you cannot obtain
the Maxwell equations, that's bad luck for your ether theory. Not for
the Maxwell equations.

>> No, they are far from arbitrary, they have well-defined measurement
>> procedures as the definition.  This definition is usually based on the
>> actually most accurate way to measure the given thing. (That's why
>> these definition are sometimes changed, once a more accurate
>> measurement device is established.)
>>
>> Once you don't have a new measurement device for whatever which is
>> more accurate than all known such devices, you have no base for
>> proposing a change of any of the definitions of those units.
>
> The point is: one can define the concept of charge in a way that
> explains what it actually is without changing the results of the
> measurements that have been performed to establish it's value.

Such a "concept of a charge" may be part of your ether theory. No
problem.  But if it appears that this concept of a charge is in
conflict with the Maxwell equations, that's bad luck for this concept,
and it has to be thrown away together with the corresponding ether
theory.  And you have to try something else.

You are NOT free to change equations for well-defined observables like
E and B which have been well-tested.  EXCEPT if you are able to show
that in some limit these equations will be recovered.

>> No. The units of measurement for E and B must match the actual most
>> accurate measurement procedures for E and B, and nothing else. And I
>> would not recommend you to propose any changes.
>>
>> If your ether theory contains some fields E', B' which you, for
>> whatever reasons, want to add, then you have to introduce constants E
>> = c_e E'. B = c_B B' with the appropriate units.  These are your
>> ether-theoretical constructions.  E and B remain what they are, and
>> the SI defitions of their units remain valid too.  They make sense.
>
> I think I've made quite a step in that direction with the definitions
> proposed above.

I'm not sure. I have yet to wait for your acknowledging that curl E =
0 is dead.

>> This is your hypothesis. Unfortunately, the "transverse" part seems
>> sufficient to test all the terms used in the Mawell equations for E
>> and B.  The proposal you have made simply destroys the equations
>> completely, without leaving the "transverse part" unchanged.
>
> Not really. Conceptually, the E and B fields now propagate trough a
> medium with well defined properties and dynamics, whereby the
> Helmholtz decompositon forms the fundamental theorem that defines the
> dynamics of the medium via the definition of potentials and velocity
> fields, that can be related to the existing fields by actually
> defining what charge is and then it also follows from a units of
> measurement analysis what current is and what unit of measurement it
> should have,

Whatever. If it follows from your considerations that curl E = 0 even
if B changes, your theory is dead.

It is, of course, also dead because usual light is not described, but
we should concentrate on Faraday's experiment, either with wires with
high enough resistance so that the resulting distortion of the E field
is not decisive, or conceptually clearer (even if much less accurate)
with charged kork balls and no wires at all used for measuring the E
field.

>>> This relation is actually incorrect. See Elmore.
>>
>> ???????????
>
> Ok, that was a bit vague. He reported his E-field has a longitudinal
> component, while his B field is transverse. I included the relevant
> quote in an earlier mail. But I think my conclusion was a bit too
> fast, would have to check better before I can make this claim. It is
> clear though that Maxwell's equations break down in the analysis of
> his wave and workarounds are needed.

I doubt. Don't forget that I have questioned your idea that E and B
fields have to be orthogonal. That's for waves, not for static fields
where E and B don't influence each other.

>> Please don't introduce quantum things into the discussion, we consider
>> now a quite classical situation, namely Faraday's experiment.
>
> Ok, let's leave that for later. But obviously, we need to be able to
> explain that in a classical way also, otherwise we have a problem
> since we're defining gauge freedom away....

Gauge freedom appears in the classical equations too, thus, it is
unrelated to quantum effects.

>> No. You can create, with static charges, quite arbitrary electric
>> forces (with the potential you like). Then you can put permanent
>> magnets into the situation. Also quite arbitrary. The result will be
>> static fields E and B, and they will not be perpendicular. They are
>> not connected at all as long as they don't change.
>>
>
> Ok, now let's replace the permanent magnet with an electromagnet and
> we start with a DC current.
>
> Same situation.
>
> Now we start changing the current, but slowly, say 1 Hz, or 0.1 Hz, or 0.01
> Hz.
>
> Now the B field is changing. What happens to the E-field?
>
> All of a sudden perpendicular?

The original E-field defined by the localized charges does not go
away. The changing B field leads to some E field, which is orthogonal.
The resulting field is the sum of both. This will be hardly
orthogonal.

And, similar to curl E =/= 0, it is sufficient to have one situation
where they are not orthogonal to be sure that this is not a general
law.