Ken S. Tucker wrote:


You're attempting to *disintegrate* Maxwell's equations into
"(Coulomb and magnetic)" as separate entities. However Einstein's
1905 SR paper explained why that shouldn't be done. I think you're
taking a mechanistic and reductionist view of physics that pre-dated
Maxwell, relativity and QT. My interpretation of your statement
suggest you're questioning basic Maxwell's Equations.

1. I don't give much credence to Maxwell's theory. Because it does not
take into account quantum effects. It represents light as continuos
wave, while in fact light consists of particles - photons as was
demonstrated in another Einstein's 1905 paper. Maxwells theory has
troubles with describing the "radiation reaction" effects which can be
traced to quantum-mechanical origin. I have much higher respect for
QED.


In Planck's quantum formula E = hf, what would the
frequency "f" represent if not an EM wave?

In my view f represents the frequency of the wavefunction of one
photon. Maxwell's theory cannot represent one photon at all.
It would be wrong to represent one photon by an electromagnetic plane wave.
Such a representation would mean that one photon can blacken entire
surface of a photographic plate, when in reality it makes one small
blackened dot.



2. I obtain Coulomb and magnetic potentials by applying "dressing"
transformation to the QED Hamiltonian. These potentials are given
by eq. (12.26) and (12.28), respectively in my book. You can also
notice the presence of other potentials there, like contact,
spin-orbit spin-spin, etc. Maxwell's theory has no clue about these
additional interactions, but they are very important if you want
to calculate the spectrum of the hydrogen atom in agreement with
experiment (see subsection 12.2.4).


SR and GR are based on and compatible with Maxwell's Eqs,
and they provide refinements to the spectrum.

I haven't seen much help from GR in calculations of
fine and hyperfine spectrum of hydrogen atom or Lamb's shifts.



3. You are quite right that Coulomb and magnetic potentials must
be present simultaneously, otherwise the relativistic invariance
of theory would be violated. This fact is also reflected in my approach.
I.e., if you omit magnetic potential (or any other part of the Breit
Hamiltonian in subsection (12.2.3)), the Poincare commutation relations
will no longer be valid with the accuracy of c^{-2}. So the theory
will be not relativistically invariant.


Agreed....


Not necessarily, we can take A and B as two specks of dust with equal
charges separated by 1 cm, so all quantum efects are out of the picture.
Then we can concentrate photons in a narrow laser beam, much smaller
than 1 cm, and shoot at particle A, so particle B is not directly
affected by the photons.



Interesting, Am I correct to interpret what you said as...

So A receives a momentum impulse from the high frquency laser and
that varies - in turn - the EM potential of A relative to B with
the expected result of a lower frequency emission from the system
A coupled to B?

The main question is not about the emission of photons from this system
(you can find discussion of bremsstrahlung emission in this case in
subsection 12.3.4; see also fig. 12.3). The main question is about the
time elapsed from the moment when laser beam hits particle A and the
moment when particle B reacts to this event. I claim that this time is
zero. You say, that this time is R_{AB}/c. So far, experiment was
silent.


The ground rules for your gedanken are conceptually deficient,
simply because you are localizing the "particles" to some fixed
position, it might be easier to substitute the archaic word
"particle" with "dent" conceptually. In QM the "dent" is a
probability field, in GR it's an energy density field, I think
the're both the same thing, same units, same curves.

Now can you define a dent in spacetime using a single point?
No, so rephrase your gedanken using "dents". Then we can
examine it from either QM or GR as you prefer, all the same
to me.

I don't find anything "archaic" in the word "particle". Particles
are produced and measured in experimental laboratories. They have
well defined masses, spins, charges, lifetimes, etc. I don't know what's
the meaning of "dent" is. Probability field in QM is just the
manifestation that particles do not obey classical rules of Newtonian
mechanics. They do not follow well-defined trajectories. There is
an element of unpredictability in their movement. But still, they
remain particles and when they hit a photographic plate or
bubble chamber they leave quite distinct localized traces there.

If you don't want to deal with quantum aspects of particle behavior,
that's fine. I agree that classical picture of trajectories is
much simpler. For this reason,
in my gedanken experiment, I intentionally avoided word "particle"
and used "specks of dust" - small but still macroscopic objects,
for which quantum effects can be neglected. They move along well-defined
trajectories and can be "localized to some fixed position".
I predict that electromagnetic interaction between such specks of dust
is instantaneous that can be measured in experiment.



Einstein's SR was based on interacting systems, ie the
"Electrodynamics of Moving Bodies", have you studied his paper?

Yes, Sir. Einstein's theory is based on two postulates:
1) principle of relativity
and
2) invariance of the speed of light.

I agree with both. To be consistent with postulate 2) Einstein starts
his analysis from considering event with freely propagating light
pulses. He derives all famous results, like time dilation, length
contraction, and Lorentz transformations, etc. I have no doubt,
these derivations are correct, and postulate 2) is of great help there.

In part 2) of his paper Einstein switches to transformations of
the electric and magnetic fields. This part has no appeal to me,
because, as I said, Maxwell's theory is approximate.

I am interested in boost transformations of trajectories of two
interacting particles. Although Einstein does not consider
this case in his paper, his belief in universality of Lorentz
transformations allows me to conclude that he thinks that
when viewed from different frames of reference, these trajectories
woud transform exactly by Lorentz fromulas. However, there are
big questions regarding this conclusion:

1) postulate 2) does not say anything about massive particles,
let alone interactive massive particles. So, Einstein's
believe in the universality of Lorentz transformations does not
follow from his two postulates. He should have introduced this
statement as an additional postulate.


He did in GR, which have so far demonstrated no real
respect for.

Lorentz transformations are modified in GR only due to gravity.
Electromagnetic interactions have no effect on boost transformations
of observables in GR. The whole point of my approach is the effect
of electromagnetic interactions on boost transformations.
I refuse to discuss gravity.



2) There is a theorem proven by Currie, Jordan and Sudarshan,
(see subsection 12.3.1 of my book) that if trajectories of two
(or more)
particles transform to the moving frame of reference via Lorentz,
then these particles do not interact.


There are places where SR/Lorentz is weak that's why
GR was invented.

Again, from what I read in GR textbooks, the space-time curvature and
the rest of GR is needed only for gravity. You may have an
unconventional theory which derives QM from GR. However, it does not
matter for our discussion how you derive QM.
The important thing for me is that
you respect all postulates of quantum mechanics and you respect
the principle of relativity and Poincare group properties in the
absence of gravitation. Can we agree on this starting point of our
discussion? If we do agree on this starting point, then my
conclusions about the non-universality of Lorentz transformations and
action-at-a-distance follow from here by straightforward logic.

If you want to challenge my logic, be my guest.

Eugene.




So, Einstein's theory is in trouble.


The theory is not in trouble, that I can see, the mathematical
description can be improved and extended.


Eugene Stefanovich


Regards
Ken S. Tucker


PS: Don't waste time, learn GR.

"Eugene Stefanovich" wrote in message
...
|
|
| Ken S. Tucker wrote:
|
|
| You're attempting to *disintegrate* Maxwell's equations into
| "(Coulomb and magnetic)" as separate entities. However Einstein's
| 1905 SR paper explained why that shouldn't be done. I think you're
| taking a mechanistic and reductionist view of physics that pre-dated
| Maxwell, relativity and QT. My interpretation of your statement
| suggest you're questioning basic Maxwell's Equations.
|
| 1. I don't give much credence to Maxwell's theory. Because it does not
| take into account quantum effects. It represents light as continuos
| wave, while in fact light consists of particles - photons as was
| demonstrated in another Einstein's 1905 paper. Maxwells theory has
| troubles with describing the "radiation reaction" effects which can be
| traced to quantum-mechanical origin. I have much higher respect for
| QED.
|
|
| In Planck's quantum formula E = hf, what would the
| frequency "f" represent if not an EM wave?
|
| In my view f represents the frequency of the wavefunction of one
| photon. Maxwell's theory cannot represent one photon at all.
| It would be wrong to represent one photon by an electromagnetic plane
wave.
| Such a representation would mean that one photon can blacken entire
| surface of a photographic plate, when in reality it makes one small
| blackened dot.

Maxwell's theory can represent one photon by the simple introduction of hbar
and 4pi. I have done it. See,

http://vacuum-physics.com/QVC/quantum_vacuum_charge.pdf

But yes, it is not going to be a plane wave. The Maxwell wave equations
admit to solutions other than just plane waves. Photographic plate implies
visible light or close to it. What would happen if you tried to detect a
radio wave photon by similar methods? Would it still be a small dot? Or
would it be ring-like with not much intensity in the center? According to
helix antenna design, they seem to be the latter case.

FrediFizzx

FrediFizzx wrote:
"Eugene Stefanovich" wrote in message
...
|
|
| Ken S. Tucker wrote:
|
|
| You're attempting to *disintegrate* Maxwell's equations into
| "(Coulomb and magnetic)" as separate entities. However Einstein's
| 1905 SR paper explained why that shouldn't be done. I think you're
| taking a mechanistic and reductionist view of physics that pre-dated
| Maxwell, relativity and QT. My interpretation of your statement
| suggest you're questioning basic Maxwell's Equations.
|
| 1. I don't give much credence to Maxwell's theory. Because it does not
| take into account quantum effects. It represents light as continuos
| wave, while in fact light consists of particles - photons as was
| demonstrated in another Einstein's 1905 paper. Maxwells theory has
| troubles with describing the "radiation reaction" effects which can be
| traced to quantum-mechanical origin. I have much higher respect for
| QED.
|
|
| In Planck's quantum formula E = hf, what would the
| frequency "f" represent if not an EM wave?
|
| In my view f represents the frequency of the wavefunction of one
| photon. Maxwell's theory cannot represent one photon at all.
| It would be wrong to represent one photon by an electromagnetic plane
wave.
| Such a representation would mean that one photon can blacken entire
| surface of a photographic plate, when in reality it makes one small
| blackened dot.

Maxwell's theory can represent one photon by the simple introduction of hbar
and 4pi. I have done it. See,

http://vacuum-physics.com/QVC/quantum_vacuum_charge.pdf

But yes, it is not going to be a plane wave. The Maxwell wave equations
admit to solutions other than just plane waves. Photographic plate implies
visible light or close to it. What would happen if you tried to detect a
radio wave photon by similar methods? Would it still be a small dot? Or
would it be ring-like with not much intensity in the center? According to
helix antenna design, they seem to be the latter case.

FrediFizzx

We cannot make a photographic plate for radiowaves because there is no
such sensitive material. The sensitivity must be a fraction of eV.
In the case of visible light photography, there are AgBr crystals which
experience an irreversible transformation (separation of Ag clusters)
after absorption of photons with energies of few eV.
If we can find a material which can similarly experience irreversible
transformation after absorbing 0.01 eV photons, we can make a
photographic plate for radiowaves. I would predict that even in this
case each photon will produce just a small dot.

The impotence of Maxwell's theory was demonstrated by Einstein's
explanation of photoeffect. This is easy to see on the example with
the photographic plate too. If light were an electromagnetic wave,
then we could make a photographic image of radiowave with existing
materials: just make a wave
of sufficiently high intensity. It never happens, because light
and radiowaves propagate as particles - photons. See discussion in
subsection 3.1.3 of my book.

Eugene.

Eugene Stefanovich wrote:


FrediFizzx wrote:

"Eugene Stefanovich" wrote in message
...
|
|
| Ken S. Tucker wrote:
|
|
| You're attempting to *disintegrate* Maxwell's equations into
| "(Coulomb and magnetic)" as separate entities. However Einstein's
| 1905 SR paper explained why that shouldn't be done. I think you're
| taking a mechanistic and reductionist view of physics that pre-dated
| Maxwell, relativity and QT. My interpretation of your statement
| suggest you're questioning basic Maxwell's Equations.
|
| 1. I don't give much credence to Maxwell's theory. Because it does
not
| take into account quantum effects. It represents light as continuos
| wave, while in fact light consists of particles - photons as was
| demonstrated in another Einstein's 1905 paper. Maxwells theory has
| troubles with describing the "radiation reaction" effects which
can be
| traced to quantum-mechanical origin. I have much higher respect for
| QED.
|
|
| In Planck's quantum formula E = hf, what would the
| frequency "f" represent if not an EM wave?
|
| In my view f represents the frequency of the wavefunction of one
| photon. Maxwell's theory cannot represent one photon at all.
| It would be wrong to represent one photon by an electromagnetic plane
wave.
| Such a representation would mean that one photon can blacken entire
| surface of a photographic plate, when in reality it makes one small
| blackened dot.

Maxwell's theory can represent one photon by the simple introduction
of hbar
and 4pi. I have done it. See,

http://vacuum-physics.com/QVC/quantum_vacuum_charge.pdf

But yes, it is not going to be a plane wave. The Maxwell wave equations
admit to solutions other than just plane waves. Photographic plate
implies
visible light or close to it. What would happen if you tried to detect a
radio wave photon by similar methods? Would it still be a small dot? Or
would it be ring-like with not much intensity in the center?
According to
helix antenna design, they seem to be the latter case.

FrediFizzx


We cannot make a photographic plate for radiowaves because there is no
such sensitive material. The sensitivity must be a fraction of eV.
In the case of visible light photography, there are AgBr crystals which
experience an irreversible transformation (separation of Ag clusters)
after absorption of photons with energies of few eV.
If we can find a material which can similarly experience irreversible
transformation after absorbing 0.01 eV photons, we can make a
photographic plate for radiowaves. I would predict that even in this
case each photon will produce just a small dot.

The impotence of Maxwell's theory was demonstrated by Einstein's
explanation of photoeffect. This is easy to see on the example with
the photographic plate too. If light were an electromagnetic wave,
then we could make a photographic image of radiowave with existing
materials: just make a wave
of sufficiently high intensity. It never happens, because light
and radiowaves propagate as particles - photons. See discussion in
subsection 3.1.3 of my book.

Eugene.

Suppose that the radio source is moving toward the emulsion at
relativistic speeds, i.e., so that the frequency is blue shifted into
the visible spectrum wrt the emulsion. You are supposing that little
dots appear on the emulsion, and I don't disagree. OTOH, wrt the source
it is the emulsion that is in motion, and wrt the source the wave can be
measured in the space surrounding the emulsion, and thus the wave energy
is indeed smeared out over a spherical plane. How do you suppose that
these two results follow simultaneously? The only logical conclusions
that can be derived a
1) the emulsion is frequency sensitive
2) the molecules of the emulsion are independently sensitive to incident
waves, i.e. some mechanism is coupling certain of the molecules to the
incident wave.
3) the dots are the result of chance interactions between the wave and
the phases of the electron orbits in the molecules.

IOW, the population is pumped by thermal energy, and constructive nodes
result in the absorption of energy from the wave. Simultaneously
other molecules are losing energy in their interaction with the wave.
This is simple radio theory, the difference is only in the fact that you
are viewing molecular electrons as objectively different particles than
free electrons (such as those oscillating within a dipole antenna).

It's all crap, force fit to the data.
However, yours seems to be an improved version of the crap simply
because you've concluded correctly that the electromagnetic force is
instantaneous. I derive the same necessary result in the paper linked
below.
Though screening is nothing new to plasma physicists, and though it is
sufficient to account for observed delays, it is somehow supposed as
nonexistent and/or impertinent to the the remainder of the interactions
in the universe, or so the texts would lead you to believe. The most
fundamental flaw in all of modern physics is redundancy ad infinitum.

The further down one goes, the simpler the laws and behaviors should
become. This seems a logical necessity to me, that is, taking the whole
as the sum of its parts. Excessive theories are the symptom of
misguided attempts to formulate general laws, and of lack of
communication and cooperation between the various fields of study. Just
as two sets of laws aren't required to govern the emission/absorption of
em energy by antennae and molecules respectively, neither are two
theories of em delays required to explain plasma waves and em waves.

Richard Perry
http://www.cswnet.com/~rper/Electromagnetism.html
Invariant wrt all transformation theories, the latter becoming a matter
of adjustments to the "measured" values of the system parameters rather
than laws pertaining to the invariant parameters of the system per se.

RP wrote:


Eugene Stefanovich wrote:



FrediFizzx wrote:

"Eugene Stefanovich" wrote in message
...
|
|
| Ken S. Tucker wrote:
|
|
| You're attempting to *disintegrate* Maxwell's equations into
| "(Coulomb and magnetic)" as separate entities. However Einstein's
| 1905 SR paper explained why that shouldn't be done. I think you're
| taking a mechanistic and reductionist view of physics that
pre-dated
| Maxwell, relativity and QT. My interpretation of your statement
| suggest you're questioning basic Maxwell's Equations.
|
| 1. I don't give much credence to Maxwell's theory. Because it
does not
| take into account quantum effects. It represents light as continuos
| wave, while in fact light consists of particles - photons as was
| demonstrated in another Einstein's 1905 paper. Maxwells theory has
| troubles with describing the "radiation reaction" effects which
can be
| traced to quantum-mechanical origin. I have much higher respect for
| QED.
|
|
| In Planck's quantum formula E = hf, what would the
| frequency "f" represent if not an EM wave?
|
| In my view f represents the frequency of the wavefunction of one
| photon. Maxwell's theory cannot represent one photon at all.
| It would be wrong to represent one photon by an electromagnetic plane
wave.
| Such a representation would mean that one photon can blacken entire
| surface of a photographic plate, when in reality it makes one small
| blackened dot.

Maxwell's theory can represent one photon by the simple introduction
of hbar
and 4pi. I have done it. See,

http://vacuum-physics.com/QVC/quantum_vacuum_charge.pdf

But yes, it is not going to be a plane wave. The Maxwell wave equations
admit to solutions other than just plane waves. Photographic plate
implies
visible light or close to it. What would happen if you tried to
detect a
radio wave photon by similar methods? Would it still be a small
dot? Or
would it be ring-like with not much intensity in the center?
According to
helix antenna design, they seem to be the latter case.

FrediFizzx



We cannot make a photographic plate for radiowaves because there is no
such sensitive material. The sensitivity must be a fraction of eV.
In the case of visible light photography, there are AgBr crystals which
experience an irreversible transformation (separation of Ag clusters)
after absorption of photons with energies of few eV.
If we can find a material which can similarly experience irreversible
transformation after absorbing 0.01 eV photons, we can make a
photographic plate for radiowaves. I would predict that even in this
case each photon will produce just a small dot.

The impotence of Maxwell's theory was demonstrated by Einstein's
explanation of photoeffect. This is easy to see on the example with
the photographic plate too. If light were an electromagnetic wave,
then we could make a photographic image of radiowave with existing
materials: just make a wave
of sufficiently high intensity. It never happens, because light
and radiowaves propagate as particles - photons. See discussion in
subsection 3.1.3 of my book.

Eugene.


Suppose that the radio source is moving toward the emulsion at
relativistic speeds, i.e., so that the frequency is blue shifted into
the visible spectrum wrt the emulsion. You are supposing that little
dots appear on the emulsion, and I don't disagree. OTOH, wrt the source
it is the emulsion that is in motion, and wrt the source the wave can be
measured in the space surrounding the emulsion, and thus the wave energy
is indeed smeared out over a spherical plane. How do you suppose that
these two results follow simultaneously? The only logical conclusions
that can be derived a
1) the emulsion is frequency sensitive
2) the molecules of the emulsion are independently sensitive to incident
waves, i.e. some mechanism is coupling certain of the molecules to the
incident wave.
3) the dots are the result of chance interactions between the wave and
the phases of the electron orbits in the molecules.

I don't quite understand your point here. Of course, the relative energy
of motion of emulsion atoms and the photon is relevant to the formation
of the image. If you shoot at a target with a bullet you make a hole.
If you place the bullet at rest and hit it with supersonic target you
get the same hole.


IOW, the population is pumped by thermal energy, and constructive nodes
result in the absorption of energy from the wave. Simultaneously other
molecules are losing energy in their interaction with the wave. This is
simple radio theory, the difference is only in the fact that you are
viewing molecular electrons as objectively different particles than free
electrons (such as those oscillating within a dipole antenna).

It's all crap, force fit to the data.
However, yours seems to be an improved version of the crap simply
because you've concluded correctly that the electromagnetic force is
instantaneous. I derive the same necessary result in the paper linked
below.
Though screening is nothing new to plasma physicists, and though it is
sufficient to account for observed delays, it is somehow supposed as
nonexistent and/or impertinent to the the remainder of the interactions
in the universe, or so the texts would lead you to believe. The most
fundamental flaw in all of modern physics is redundancy ad infinitum.

The further down one goes, the simpler the laws and behaviors should
become. This seems a logical necessity to me, that is, taking the whole
as the sum of its parts. Excessive theories are the symptom of
misguided attempts to formulate general laws, and of lack of
communication and cooperation between the various fields of study. Just
as two sets of laws aren't required to govern the emission/absorption of
em energy by antennae and molecules respectively, neither are two
theories of em delays required to explain plasma waves and em waves.

Richard Perry
http://www.cswnet.com/~rper/Electromagnetism.html
Invariant wrt all transformation theories, the latter becoming a matter
of adjustments to the "measured" values of the system parameters rather
than laws pertaining to the invariant parameters of the system per se.


Sorry, I hardly understand what you are saying.

Eugene.

Eugene Stefanovich wrote in message ...
Ken S. Tucker wrote:

1. I don't give much credence to Maxwell's theory. Because it does not
take into account quantum effects. It represents light as continuos
wave, while in fact light consists of particles - photons as was
demonstrated in another Einstein's 1905 paper. Maxwells theory has
troubles with describing the "radiation reaction" effects which can be
traced to quantum-mechanical origin. I have much higher respect for
QED.

In Planck's quantum formula E = hf, what would the
frequency "f" represent if not an EM wave?

In my view f represents the frequency of the wavefunction of one
photon. Maxwell's theory cannot represent one photon at all.
It would be wrong to represent one photon by an electromagnetic plane wave.
Such a representation would mean that one photon can blacken entire
surface of a photographic plate, when in reality it makes one small
blackened dot.

A photon must possess a certain *invariant dynamic structure*.
Imagine a visible photon. Relative to a FoR (Robert) receding
from it's source the photon is Doppler shifted into the radio
frequency range, but relative to an approaching FoR (Allan) it's
shifted to a gamma ray. Allan finds the gamma ray consists of 2
charges, by mutating it to a positron and an electron. Robert
has a sensitive radio antenna and measures the photon to have
a varying E and B field, depending on how he polarizes his antenna.

Robert and Allan are observing the *same* photon, so the
differences in their measurements are entirely *relativistic*.

Is that agreeable?

2. I obtain Coulomb and magnetic potentials by applying "dressing"
transformation to the QED Hamiltonian. These potentials are given
by eq. (12.26) and (12.28), respectively in my book. You can also
notice the presence of other potentials there, like contact,
spin-orbit spin-spin, etc. Maxwell's theory has no clue about these
additional interactions, but they are very important if you want
to calculate the spectrum of the hydrogen atom in agreement with
experiment (see subsection 12.2.4).

SR and GR are based on and compatible with Maxwell's Eqs,
and they provide refinements to the spectrum.

I haven't seen much help from GR in calculations of
fine and hyperfine spectrum of hydrogen atom or Lamb's shifts.

Steve Bell, is an occasional contributor to this NG, and
has done some very detailed analysis on orbitals based on
Kerr metrics, I would recommend you email him for commentary
on your new book.
[...]

The ground rules for your gedanken are conceptually deficient,
simply because you are localizing the "particles" to some fixed
position, it might be easier to substitute the archaic word
"particle" with "dent" conceptually. In QM the "dent" is a
probability field, in GR it's an energy density field, I think
the're both the same thing, same units, same curves.

Now can you define a dent in spacetime using a single point?
No, so rephrase your gedanken using "dents". Then we can
examine it from either QM or GR as you prefer, all the same
to me.

I don't find anything "archaic" in the word "particle". Particles
are produced and measured in experimental laboratories. They have
well defined masses, spins, charges, lifetimes, etc. I don't know what's
the meaning of "dent" is. Probability field in QM is just the
manifestation that particles do not obey classical rules of Newtonian
mechanics. They do not follow well-defined trajectories. There is
an element of unpredictability in their movement. But still, they
remain particles and when they hit a photographic plate or
bubble chamber they leave quite distinct localized traces there.

See the photon experiment above, Robert can't use a photograph,
IOW's how would you characterize the location of a radio wave?

If you don't want to deal with quantum aspects of particle behavior,
that's fine. I agree that classical picture of trajectories is
much simpler. For this reason,
in my gedanken experiment, I intentionally avoided word "particle"
and used "specks of dust" - small but still macroscopic objects,
for which quantum effects can be neglected. They move along well-defined
trajectories and can be "localized to some fixed position".

ok

I predict that electromagnetic interaction between such specks of dust
is instantaneous that can be measured in experiment.

That's a clear statement, you're predicting an infinite speed of
"electromagnetic interaction" aka "light" because that is what
your theory needs, is that correct?

He did in GR, which have so far demonstrated no real
respect for.

Lorentz transformations are modified in GR only due to gravity.
Electromagnetic interactions have no effect on boost transformations
of observables in GR. The whole point of my approach is the effect
of electromagnetic interactions on boost transformations.
I refuse to discuss gravity.

ok

2) There is a theorem proven by Currie, Jordan and Sudarshan,
(see subsection 12.3.1 of my book) that if trajectories of two
(or more)
particles transform to the moving frame of reference via Lorentz,
then these particles do not interact.

Is that verified using tensor analysis?

There are places where SR/Lorentz is weak that's why
GR was invented.

Again, from what I read in GR textbooks, the space-time curvature and
the rest of GR is needed only for gravity.

You may have an unconventional theory which derives QM from GR.

It's a deduction I paraphased from a very solid reference, ie.
Einstein's GR Foundation 1916, did you check it?
I'm happy to help you understand it.

However, it does not matter for our discussion how you derive QM.

Sure does, it places QM on a solid foundation of principle.

The important thing for me is that
you respect all postulates of quantum mechanics and you respect
the principle of relativity and Poincare group properties in the
absence of gravitation. Can we agree on this starting point of our
discussion?

Sure, but with the rules of GR.

If we do agree on this starting point, then my
conclusions about the non-universality of Lorentz transformations and
action-at-a-distance follow from here by straightforward logic.
If you want to challenge my logic, be my guest.

Ok, how does FTL communication occur? By particle, wave, another
dimension?

Eugene.

Ken S. Tucker wrote:
Eugene Stefanovich wrote in message ...

Ken S. Tucker wrote:


1. I don't give much credence to Maxwell's theory. Because it does not
take into account quantum effects. It represents light as continuos
wave, while in fact light consists of particles - photons as was
demonstrated in another Einstein's 1905 paper. Maxwells theory has
troubles with describing the "radiation reaction" effects which can be
traced to quantum-mechanical origin. I have much higher respect for
QED.



In Planck's quantum formula E = hf, what would the
frequency "f" represent if not an EM wave?



In my view f represents the frequency of the wavefunction of one
photon. Maxwell's theory cannot represent one photon at all.
It would be wrong to represent one photon by an electromagnetic plane wave.
Such a representation would mean that one photon can blacken entire
surface of a photographic plate, when in reality it makes one small
blackened dot.


A photon must possess a certain *invariant dynamic structure*.
Imagine a visible photon. Relative to a FoR (Robert) receding
from it's source the photon is Doppler shifted into the radio
frequency range, but relative to an approaching FoR (Allan) it's
shifted to a gamma ray. Allan finds the gamma ray consists of 2
charges, by mutating it to a positron and an electron. Robert
has a sensitive radio antenna and measures the photon to have
a varying E and B field, depending on how he polarizes his antenna.

Robert and Allan are observing the *same* photon, so the
differences in their measurements are entirely *relativistic*.

Is that agreeable?

So far so good, except for the fact that one radio photon is not
enough to create detectable wave. You need to have a huge number of them
working together in order to create continuous "electromagnetic field".
Another objection is that gamma photon does not consist of two charges.
It can create an electron-positron pair when it hits some obstacle.
Otherwise I am with you.



2. I obtain Coulomb and magnetic potentials by applying "dressing"
transformation to the QED Hamiltonian. These potentials are given
by eq. (12.26) and (12.28), respectively in my book. You can also
notice the presence of other potentials there, like contact,
spin-orbit spin-spin, etc. Maxwell's theory has no clue about these
additional interactions, but they are very important if you want
to calculate the spectrum of the hydrogen atom in agreement with
experiment (see subsection 12.2.4).



SR and GR are based on and compatible with Maxwell's Eqs,
and they provide refinements to the spectrum.



I haven't seen much help from GR in calculations of
fine and hyperfine spectrum of hydrogen atom or Lamb's shifts.


Steve Bell, is an occasional contributor to this NG, and
has done some very detailed analysis on orbitals based on
Kerr metrics, I would recommend you email him for commentary
on your new book.
[...]

Thanks for the advice. Do you know his e-mail?



The ground rules for your gedanken are conceptually deficient,
simply because you are localizing the "particles" to some fixed
position, it might be easier to substitute the archaic word
"particle" with "dent" conceptually. In QM the "dent" is a
probability field, in GR it's an energy density field, I think
the're both the same thing, same units, same curves.

Now can you define a dent in spacetime using a single point?
No, so rephrase your gedanken using "dents". Then we can
examine it from either QM or GR as you prefer, all the same
to me.

I don't find anything "archaic" in the word "particle". Particles
are produced and measured in experimental laboratories. They have
well defined masses, spins, charges, lifetimes, etc. I don't know what's
the meaning of "dent" is. Probability field in QM is just the
manifestation that particles do not obey classical rules of Newtonian
mechanics. They do not follow well-defined trajectories. There is
an element of unpredictability in their movement. But still, they
remain particles and when they hit a photographic plate or
bubble chamber they leave quite distinct localized traces there.


See the photon experiment above, Robert can't use a photograph,
IOW's how would you characterize the location of a radio wave?

That's just indicates the inadequacy of our experimental devices.
Theoretically, if there existed radio-photography, then radio photons
would make tiny little dots on the "photograph". I cannot prove that,
I believe in that because I believe in unity of natu photons
of different energy have exactly the same properties except
energy. How would you describe the radio-photgraphy experiment?



If you don't want to deal with quantum aspects of particle behavior,
that's fine. I agree that classical picture of trajectories is
much simpler. For this reason,
in my gedanken experiment, I intentionally avoided word "particle"
and used "specks of dust" - small but still macroscopic objects,
for which quantum effects can be neglected. They move along well-defined
trajectories and can be "localized to some fixed position".


ok


I predict that electromagnetic interaction between such specks of dust
is instantaneous that can be measured in experiment.


That's a clear statement, you're predicting an infinite speed of
"electromagnetic interaction" aka "light" because that is what
your theory needs, is that correct?

No, light has nothing (or very little) to do with electromagnetic
interactions. Light is a flow of photons. Electromagnetic interactions
are Coulomb and magnetic (and contact and spin-orbit...) instantaneous
potentials (or I should better say "forces", so you not confuse them
with the scalar and vector potentials of Maxwell's theory) acting
between charged particles (see subsection 12.2.3).
The connection between light and electromagnetism appears in my theory
if
we take into consideration the bremsstrahlung terms in the Hamiltonian
(see Table 1 in section 12.1). In classical Maxwell's physics this
interaction has counterpart in "radiation reaction". Due to
bremsstrahlung (radiation due to acceleration or breaking), each time
a charged particle is accelerated or decelerated it emits (or absorbs)
photons.
So, when you make a call on your cell phone, the electrons in the
antenna start to move back and forth. This creates two effects in
the neighborhood. First there appear bremsstrahlung photons which
fly away with the speed of light and either go to space or hit receiving
antenna. Thanks to these photons we have radio communication.
Second, moving electrons in your antenna interact with neighboring
charges via Coulomb and magnetic forces. These forces are instantaneous.
In addition they decrease rather rapidly with the distance from the antenna.
For example, the Coulomb force decreases proportionally to the square of
the distance
(the variation in force decreases even faster). So, the effect of these
instantaneous forces is localized around your cell phone. Certainly,
you cannot use this force to communicate instantaneously with
alpha Centauri. You can use photons (or transversal radio wave,
in classical language) for communication, but, sadly, they move with
the speed of
light.



He did in GR, which have so far demonstrated no real
respect for.

Lorentz transformations are modified in GR only due to gravity.
Electromagnetic interactions have no effect on boost transformations
of observables in GR. The whole point of my approach is the effect
of electromagnetic interactions on boost transformations.
I refuse to discuss gravity.


ok


2) There is a theorem proven by Currie, Jordan and Sudarshan,
(see subsection 12.3.1 of my book) that if trajectories of two
(or more)
particles transform to the moving frame of reference via Lorentz,
then these particles do not interact.


Is that verified using tensor analysis?

The original version of the theorem (I reproduce its simplified proof in
subsection 12.3.1) uses Poisson brackets between particle observables
and generators of inertial transformations. This theorem has been proven
in 1963, and since then there were many papers with generalizations of
this theorem to include many particles, spins, etc. I know at least
half a dozen such generalizations, but I am sure there are more. Maybe
some of them use tensor analysis, I don't know. Why do you ask?




There are places where SR/Lorentz is weak that's why
GR was invented.

Again, from what I read in GR textbooks, the space-time curvature and
the rest of GR is needed only for gravity.


You may have an unconventional theory which derives QM from GR.


It's a deduction I paraphased from a very solid reference, ie.
Einstein's GR Foundation 1916, did you check it?
I'm happy to help you understand it.


However, it does not matter for our discussion how you derive QM.


Sure does, it places QM on a solid foundation of principle.

I used another foundation for QM called "quantum logic" in section 4.1.
But this is not relevant. We both agree that QM is correct. Let's move
on.



The important thing for me is that
you respect all postulates of quantum mechanics and you respect
the principle of relativity and Poincare group properties in the
absence of gravitation. Can we agree on this starting point of our
discussion?


Sure, but with the rules of GR.

Fine.




If we do agree on this starting point, then my
conclusions about the non-universality of Lorentz transformations and
action-at-a-distance follow from here by straightforward logic.
If you want to challenge my logic, be my guest.


Ok, how does FTL communication occur? By particle, wave, another
dimension?

There is just an instantaneous potential, e.g., 1/R for the Coulomb
part, which is a part of the total Hamiltonian. There is no any
mechanical carrier for this interaction. Why you think the presence
of such a carrier is necessary?



Eugene.

"Ken S. Tucker" wrote in message
om...
| Eugene Stefanovich wrote in message
...
| Ken S. Tucker wrote:
|
| 1. I don't give much credence to Maxwell's theory. Because it does not
| take into account quantum effects. It represents light as continuos
| wave, while in fact light consists of particles - photons as was
| demonstrated in another Einstein's 1905 paper. Maxwells theory has
| troubles with describing the "radiation reaction" effects which can be
| traced to quantum-mechanical origin. I have much higher respect for
| QED.
|
| In Planck's quantum formula E = hf, what would the
| frequency "f" represent if not an EM wave?
|
| In my view f represents the frequency of the wavefunction of one
| photon. Maxwell's theory cannot represent one photon at all.
| It would be wrong to represent one photon by an electromagnetic plane
wave.
| Such a representation would mean that one photon can blacken entire
| surface of a photographic plate, when in reality it makes one small
| blackened dot.
|
| A photon must possess a certain *invariant dynamic structure*.

And that "structure" is what exactly?

| Imagine a visible photon. Relative to a FoR (Robert) receding
| from it's source the photon is Doppler shifted into the radio
| frequency range, but relative to an approaching FoR (Allan) it's
| shifted to a gamma ray. Allan finds the gamma ray consists of 2
| charges, by mutating it to a positron and an electron. Robert
| has a sensitive radio antenna and measures the photon to have
| a varying E and B field, depending on how he polarizes his antenna.
|
| Robert and Allan are observing the *same* photon, so the
| differences in their measurements are entirely *relativistic*.
|
| Is that agreeable?

I would have to say that Robert and Allan aren't observing the same photon.
So not agreeable. A photon in this case is not defined properly until it is
detected (destroyed).

FrediFizzx

Eugene Stefanovich wrote in message ...
Ken S. Tucker wrote:
Eugene Stefanovich wrote in message ...
[...]
In my view f represents the frequency of the wavefunction of one
photon. Maxwell's theory cannot represent one photon at all.
It would be wrong to represent one photon by an electromagnetic plane wave.
Such a representation would mean that one photon can blacken entire
surface of a photographic plate, when in reality it makes one small
blackened dot.

A photon must possess a certain *invariant dynamic structure*.
Imagine a visible photon. Relative to a FoR (Robert) receding
from it's source the photon is Doppler shifted into the radio
frequency range, but relative to an approaching FoR (Allan) it's
shifted to a gamma ray. Allan finds the gamma ray consists of 2
charges, by mutating it to a positron and an electron. Robert
has a sensitive radio antenna and measures the photon to have
a varying E and B field, depending on how he polarizes his antenna.

Robert and Allan are observing the *same* photon, so the
differences in their measurements are entirely *relativistic*.

Is that agreeable?

So far so good, except for the fact that one radio photon is not
enough to create detectable wave.

We're digressing into a technology issue.

You need to have a huge number of them
working together in order to create continuous "electromagnetic field".

"huge" what is the minimum number? How do they "work together"?

Another objection is that gamma photon does not consist of two charges.
It can create an electron-positron pair when it hits some obstacle.

Not quite, pair production does require a field, but doesn't use the
photoelectric effect.

Otherwise I am with you.

We're quite apart.
You've implied a single "radio" photon cannot be measured, and more
over this single photon has no E or B field!

I haven't seen much help from GR in calculations of
fine and hyperfine spectrum of hydrogen atom or Lamb's shifts.

Steve Bell, is an occasional contributor to this NG, and
has done some very detailed analysis on orbitals based on
Kerr metrics, I would recommend you email him for commentary
on your new book.
[...]

Thanks for the advice. Do you know his e-mail?

Not off hand, Google search his name in sci.physics - relativity.
He has some interesting articles he's posted, and a book, we haven't
emailed for some time now.

See the photon experiment above, Robert can't use a photograph,
IOW's how would you characterize the location of a radio wave?

That's just indicates the inadequacy of our experimental devices.
Theoretically, if there existed radio-photography, then radio photons
would make tiny little dots on the "photograph". I cannot prove that,
I believe in that because I believe in unity of natu photons
of different energy have exactly the same properties except
energy. How would you describe the radio-photgraphy experiment?

This is a photon absorption problem, and we fall back on to
wavelength to determine that, even when talking about the resolution
and penetration ability of photons at various frequencies.
I would guess that a 1 cm wavelength photon would leave a 1 cm line.
[...]

That's a clear statement, you're predicting an infinite speed of
"electromagnetic interaction" aka "light" because that is what
your theory needs, is that correct?

No, light has nothing (or very little) to do with electromagnetic
interactions. Light is a flow of photons. Electromagnetic interactions
are Coulomb and magnetic (and contact and spin-orbit...) instantaneous
potentials (or I should better say "forces", so you not confuse them
with the scalar and vector potentials of Maxwell's theory) acting
between charged particles (see subsection 12.2.3).

The connection between light and electromagnetism appears in my theory
if
we take into consideration the bremsstrahlung terms in the Hamiltonian
(see Table 1 in section 12.1). In classical Maxwell's physics this
interaction has counterpart in "radiation reaction". Due to
bremsstrahlung (radiation due to acceleration or breaking), each time
a charged particle is accelerated or decelerated it emits (or absorbs)
photons.
So, when you make a call on your cell phone, the electrons in the
antenna start to move back and forth. This creates two effects in
the neighborhood. First there appear bremsstrahlung photons which
fly away with the speed of light and either go to space or hit receiving
antenna. Thanks to these photons we have radio communication.
Second, moving electrons in your antenna interact with neighboring
charges via Coulomb and magnetic forces. These forces are instantaneous.
In addition they decrease rather rapidly with the distance from the antenna.
For example, the Coulomb force decreases proportionally to the square of
the distance
(the variation in force decreases even faster). So, the effect of these
instantaneous forces is localized around your cell phone. Certainly,
you cannot use this force to communicate instantaneously with
alpha Centauri. You can use photons (or transversal radio wave,
in classical language) for communication, but, sadly, they move with
the speed of
light.

[...]

Is that verified using tensor analysis?

The original version of the theorem (I reproduce its simplified proof in
subsection 12.3.1) uses Poisson brackets between particle observables
and generators of inertial transformations. This theorem has been proven
in 1963, and since then there were many papers with generalizations of
this theorem to include many particles, spins, etc. I know at least
half a dozen such generalizations, but I am sure there are more. Maybe
some of them use tensor analysis, I don't know. Why do you ask?

Tensor analysis filters out CS artifacts and leaves invariant
relations.

If we do agree on this starting point, then my
conclusions about the non-universality of Lorentz transformations and
action-at-a-distance follow from here by straightforward logic.
If you want to challenge my logic, be my guest.


Ok, how does FTL communication occur? By particle, wave, another
dimension?

There is just an instantaneous potential, e.g., 1/R for the Coulomb
part, which is a part of the total Hamiltonian. There is no any
mechanical carrier for this interaction. Why you think the presence
of such a carrier is necessary?

Above you state,
"instantaneous potentials (or I should better say "forces","
Well we know from experimental evidence that energy (force)
conveyance cannot exceed the local c. Basically how do you get
around that and still retain SR?

Ken

"Eugene Stefanovich" wrote in message
...
|
|
| Ken S. Tucker wrote:
| Eugene Stefanovich wrote in message
...
|
| Ken S. Tucker wrote:
|
|
| 1. I don't give much credence to Maxwell's theory. Because it does not
| take into account quantum effects. It represents light as continuos
| wave, while in fact light consists of particles - photons as was
| demonstrated in another Einstein's 1905 paper. Maxwells theory has
| troubles with describing the "radiation reaction" effects which can be
| traced to quantum-mechanical origin. I have much higher respect for
| QED.
|
|
|
| In Planck's quantum formula E = hf, what would the
| frequency "f" represent if not an EM wave?
|
|
|
| In my view f represents the frequency of the wavefunction of one
| photon. Maxwell's theory cannot represent one photon at all.
| It would be wrong to represent one photon by an electromagnetic plane
wave.
| Such a representation would mean that one photon can blacken entire
| surface of a photographic plate, when in reality it makes one small
| blackened dot.
|
|
| A photon must possess a certain *invariant dynamic structure*.
| Imagine a visible photon. Relative to a FoR (Robert) receding
| from it's source the photon is Doppler shifted into the radio
| frequency range, but relative to an approaching FoR (Allan) it's
| shifted to a gamma ray. Allan finds the gamma ray consists of 2
| charges, by mutating it to a positron and an electron. Robert
| has a sensitive radio antenna and measures the photon to have
| a varying E and B field, depending on how he polarizes his antenna.
|
| Robert and Allan are observing the *same* photon, so the
| differences in their measurements are entirely *relativistic*.
|
| Is that agreeable?
|
| So far so good, except for the fact that one radio photon is not
| enough to create detectable wave. You need to have a huge number of them
| working together in order to create continuous "electromagnetic field".
| Another objection is that gamma photon does not consist of two charges.
| It can create an electron-positron pair when it hits some obstacle.
| Otherwise I am with you.

No Eugene, it is not "so far so good". You can't specify that the photon is
a visible light photon beforehand (before detection). This is a false
"gedanken". And it is also impossible for Robert and Allan to observe the
same photon because they are in different "frames". The photon Robert
detects is not the same photon that Allan detects.

FrediFizzx