"FrediFizzx" wrote in message ...
"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?

Freddi, I'm concerned about getting off Eugene's topic, (he's the
OP) specifically how Maxwell's Theory relates to a photon.

| 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

Technically correct, but in the world of gedanken a laser can
be assumed to create many identical photons. Any of those
photons are relatively a gamma ray or a radio wave.
Ken

"Ken S. Tucker" wrote in message
om...
| "FrediFizzx" wrote in message
...
| "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?
|
| Freddi, I'm concerned about getting off Eugene's topic, (he's the
| OP) specifically how Maxwell's Theory relates to a photon.

Actually, this goes to the heart of how Maxwell relates to a photon. A
photon has a helicity of plus or minus one. That is one part that would be
invariant.

| | 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
|
| Technically correct, but in the world of gedanken a laser can
| be assumed to create many identical photons. Any of those
| photons are relatively a gamma ray or a radio wave.

When you change frames, they are not going to be the same photons that the
laser created. In a "picture" with a relativistic medium, all that counts
is the point at which the photon is being destroyed. A photon created at
point A is not the same "entity" at B when detected. In my quantum vacuum
charge model, all gauge bosons are composites of virtual fermions that are
bound. Photons do not have intrinsic invariant EM properties.

FrediFizzx

Ken S. Tucker wrote:
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?

You can get a good estimate by dividing the total energy pumped into
the wave by hv, where v is wave's frequency.

How do they "work together"?

I don't know yet. I haven't thought about that.
This would require developing of theory of photon emission by antennas.
I am not prepared to do that now.

Look, Einstein clearly proved that light consists of photons.
I think nobody can argue that gamma rays are not particles.
I don't see a reason to believe that radio-emission is in any way
different. Simply the energy/frequency of radio photons is so
small that their corpuscular properties are not obvious. They can
be more easily described in the wave picture, but this does not mean
that such photons do not exist.



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.

By "obstacle" I meant any object (e.g., atomic nucleus) which in your
language produces a field. I didn't tell anything about the
photoelectric effect. It has nothing to do with pair creation.


Otherwise I am with you.


We're quite apart.
You've implied a single "radio" photon cannot be measured,

That's not what I was said. I said it can be measured in principle,
but we do not have adequate measuring apparatuses (like photographic
plates for visible photons) at our disposal.

and more
over this single photon has no E or B field!

Yes, there are no E and B fields in
my approach. The fields are not measurable by themselves. Each
time you claim you measured a field you actually measure some
accelerations of
particles. You say that the particles have accelerated because there
were E and B fields created by other particle. I say that particles
have accelerated because there was direct action-at-a-distance from
other particles. So, I eliminate the E and B fields from the picture.
There are particles and instantaneous interactions between them.
That's it.



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.
[...]

The wavelength of visible light is (roughly) 1 micrometer. That's the
"size" of the photon according to you.
The size of atom is (roughly) 0.1 nanometer. That's, 4 orders
of magnitude difference. How would you explain the visible light
emission and absorption by individual atoms?



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 you think there is a mistake in the proof of the CJS theorem,
I am happy to discuss it with you. We can start our discussion with
subsection 12.3.1 of my book, if you don't mind.



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.

What experimental evidence? In this thread and in other threads,
I asked many times to show me an experiment
which directly measures the speed of propagation of interactions.
Nobody was able to do that so far.

Basically how do you get
around that and still retain SR?

First, I do not retain SR. SR means universal linear (tensor) Lorentz
transformations for all observables in all physical systems.
I do not retain that. SR means universal Minkowski background
space-time. I am not cool with that either.

The points where I do agree with SR are
1) relativity postulate
2) Poincare group properties of inertial transformations.

I said many times that in my approach instantaneous propagation of
interactions does not contradict causality (as in SR). There is no
"grandfather paradox" in my approach.



Ken

Eugene

FrediFizzx wrote:
"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


Ken wrote "the *same* photon" not "the same photon". I think he
understands that once detected by one observer a photon cannot be
detected by others. But I don't think there is an issue here. We can
consider two *identical* photons: one for Robert and one for Allen.

Eugene.

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

Ken S. Tucker wrote:

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


"huge" what is the minimum number?

You can get a good estimate by dividing the total energy pumped into
the wave by hv, where v is wave's frequency.

Somehow you intend to create a formation of photons to
possess chararcteristics an individual photon does not
have. Indeed your theory depends on that structure.

How do they "work together"?

I don't know yet. I haven't thought about that.
This would require developing of theory of photon emission by antennas.
I am not prepared to do that now.

Again, your theory depends on that. How is "induction"
(radio communication) to be explained without Maxwell's Equations?

Look, Einstein clearly proved that light consists of photons.
I think nobody can argue that gamma rays are not particles.
I don't see a reason to believe that radio-emission is in any way
different. Simply the energy/frequency of radio photons is so
small that their corpuscular properties are not obvious. They can
be more easily described in the wave picture, but this does not mean
that such photons do not exist.

Your singing to the choir, we established that.

[snip ok]

Otherwise I am with you.


We're quite apart.
You've implied a single "radio" photon cannot be measured,

That's not what I was said. I said it can be measured in principle,
but we do not have adequate measuring apparatuses (like photographic
plates for visible photons) at our disposal.

and more
over this single photon has no E or B field!

Yes, there are no E and B fields in
my approach. The fields are not measurable by themselves. Each
time you claim you measured a field you actually measure some
accelerations of
particles.
You say that the particles have accelerated because there
were E and B fields created by other particle. I say that particles
have accelerated because there was direct action-at-a-distance from
other particles. So, I eliminate the E and B fields from the picture.
There are particles and instantaneous interactions between them.
That's it.

Agreed, the E and B fields by themselves are operators,
really just imports from engineering now and possess no
independant reality. They do actualize when effecting a
charge to produce a measurable current of voltage increment.

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.
[...]

The wavelength of visible light is (roughly) 1 micrometer. That's the
"size" of the photon according to you.
The size of atom is (roughly) 0.1 nanometer. That's, 4 orders
of magnitude difference. How would you explain the visible light
emission and absorption by individual atoms?

That's a digression, ie. dirty water is opague has nothing
to do with atomic structure, sigh.
[...]

Tensor analysis filters out CS artifacts and leaves invariant
relations.

If you think there is a mistake in the proof of the CJS theorem,
I am happy to discuss it with you. We can start our discussion with
subsection 12.3.1 of my book, if you don't mind.

Your Eq.(9) is not covariant. It's contingent upon the author
to provide any new "physical laws" with a covariant basis,
otherwise you're extrapolating generalities from narrow specifics.
That is a poor specification for a Professional Theoretician
to construct a theory on. By analogy, you may find a board over
a ditch is a good bridge, but that doesn't scale when building
the Golden Gate. Engineers must obey general facts, as you must
do if you intend to turn professional.
[...]

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.

What experimental evidence? In this thread and in other threads,
I asked many times to show me an experiment
which directly measures the speed of propagation of interactions.
Nobody was able to do that so far.

Yes, you're ignoring E' = E / sqrt(1 - v2/c2),

Basically how do you get
around that and still retain SR?

First, I do not retain SR. SR means universal linear (tensor) Lorentz
transformations for all observables in all physical systems.
I do not retain that. SR means universal Minkowski background
space-time. I am not cool with that either.

Would you please specify a precise equation that doesn't
need 450 pages to explain, something like E=mc^2 that
we the people may examine. If you can't then deep-six
the book, don't waste time.

The points where I do agree with SR are
1) relativity postulate
2) Poincare group properties of inertial transformations.

You claimed not to understand that (GR) before....?

I said many times that in my approach instantaneous propagation of
interactions does not contradict causality (as in SR). There is no
"grandfather paradox" in my approach.

Eugene, becoming a theoretician is much more than gluing
a bunch of notes together. From chapter to chapter your
new book may overwhelm the reader with insignificant tactical
detail, but the strategy binding the purpose is lacking,
or merely implied. At the outset, you'll definitely need
to detail a concise strategy including a clear mathematical
definition of the new physicals law you propose and those
conventional laws you intend to revolutionize, ie an abstract.

Are you prepared to provide an abstract?
Ken

Eugene

"FrediFizzx" wrote in message ...

When you change frames, they are not going to be the same photons that the
laser created. In a "picture" with a relativistic medium, all that counts
is the point at which the photon is being destroyed. A photon created at
point A is not the same "entity" at B when detected. In my quantum vacuum
charge model, all gauge bosons are composites of virtual fermions that are
bound.

Photons do not have intrinsic invariant EM properties.
FrediFizzx

I argue they do. I think that because I underastand an
individual photon can be polarized, correct me if I'm
wrong please.
Ken

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

Ken S. Tucker wrote:

Eugene Stefanovich wrote in message ...


Ken S. Tucker wrote:


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


"huge" what is the minimum number?

You can get a good estimate by dividing the total energy pumped into
the wave by hv, where v is wave's frequency.


Somehow you intend to create a formation of photons to
possess chararcteristics an individual photon does not
have. Indeed your theory depends on that structure.

Let me set it straight: photons do not interact with each other
(actually, they do interact in QED, but this interaction is very
weak and has not been observed by experiment yet). Each photon
has a wave function whose properties (e.g., frequency) get translated
to the properties of the "electromagnetic wave" where there are
many photons.

You can obtain interference picture by shooting photons one by one
through double-slit. This experiment cannot be described by
Maxwell's equations at all. It requires quantum mechanics.
You will obtain the same interference
picture if you shoot all photons at once. Then you can imagine that
you are dealing with "electromagnetic wave" and calculate the
interference picture just as Fresnel and Yound did 200 years ago.



How do they "work together"?

I don't know yet. I haven't thought about that.
This would require developing of theory of photon emission by antennas.
I am not prepared to do that now.


Again, your theory depends on that. How is "induction"
(radio communication) to be explained without Maxwell's Equations?

Electrons in the antenna start to move back and forth under the
influence of external force. Acceleration of electrons results in
emission of bremsstrahlung photons. These photons propagate with the
speed of light. When they reach receiving antenna they kick electrons
there (Compton effect) and force them to oscillate. These oscillations
are recorded by the receiver's circuitry. In order to explain that you
don't need Maxwell's equations and fields. You just need the dressed
particle Hamiltonian with interaction terms whose examples are given
in Table 1 in chapter 12 and in section 12.2.



Look, Einstein clearly proved that light consists of photons.
I think nobody can argue that gamma rays are not particles.
I don't see a reason to believe that radio-emission is in any way
different. Simply the energy/frequency of radio photons is so
small that their corpuscular properties are not obvious. They can
be more easily described in the wave picture, but this does not mean
that such photons do not exist.


Your singing to the choir, we established that.

[snip ok]


Otherwise I am with you.


We're quite apart.
You've implied a single "radio" photon cannot be measured,

That's not what I was said. I said it can be measured in principle,
but we do not have adequate measuring apparatuses (like photographic
plates for visible photons) at our disposal.


and more
over this single photon has no E or B field!

Yes, there are no E and B fields in
my approach. The fields are not measurable by themselves. Each
time you claim you measured a field you actually measure some
accelerations of
particles.
You say that the particles have accelerated because there
were E and B fields created by other particle. I say that particles
have accelerated because there was direct action-at-a-distance from
other particles. So, I eliminate the E and B fields from the picture.
There are particles and instantaneous interactions between them.
That's it.


Agreed, the E and B fields by themselves are operators,
really just imports from engineering now and possess no
independant reality. They do actualize when effecting a
charge to produce a measurable current of voltage increment.

I do not have E and B fields even as operators in my approach.
I have just direct interparticle interactions.



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.
[...]



The wavelength of visible light is (roughly) 1 micrometer. That's the
"size" of the photon according to you.
The size of atom is (roughly) 0.1 nanometer. That's, 4 orders
of magnitude difference. How would you explain the visible light
emission and absorption by individual atoms?


That's a digression, ie. dirty water is opague has nothing
to do with atomic structure, sigh.
[...]

I don't think that's a digression. When 1 cm wavelength photon
gets absorbed by a 0.1 nanometer atom, it leaves 0.1 nanometer
"line" (in your language) not 1 cm line.



Tensor analysis filters out CS artifacts and leaves invariant
relations.

If you think there is a mistake in the proof of the CJS theorem,
I am happy to discuss it with you. We can start our discussion with
subsection 12.3.1 of my book, if you don't mind.


Your Eq.(9) is not covariant. It's contingent upon the author
to provide any new "physical laws" with a covariant basis,
otherwise you're extrapolating generalities from narrow specifics.

Are you talking about eq. (9) in chapter 12? This equation gives
approximation for the "dressed particle" interaction in the 2nd
perturbation order. It is relativistically invariant with the 2nd order
accuracy. If you take contributions from all orders, the theory
is proven to be exactly relativistically invariant (see subsection
12.1.7)


That is a poor specification for a Professional Theoretician
to construct a theory on. By analogy, you may find a board over
a ditch is a good bridge, but that doesn't scale when building
the Golden Gate. Engineers must obey general facts, as you must
do if you intend to turn professional.
[...]


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.

What experimental evidence? In this thread and in other threads,
I asked many times to show me an experiment
which directly measures the speed of propagation of interactions.
Nobody was able to do that so far.


Yes, you're ignoring E' = E / sqrt(1 - v2/c2),

You haven't answered my question, from which I conclude that you don't
know any experiment which directly measures the speed of propagation
of interactions.

I don't know how your formula relates to this question.
If you want to discuss this formula now, let's do it. You haven't
mentioned, but I presume that E and E' stand for the energy of a system
in two reference frames. This formula is exactly valid in my theory
if E and E' refer to the TOTAL energy of the system. If E and E'
refer to the energy of a part of the system (e.g., one particle in
the system of interacting particles), then this formula becomes
approximate. There are corrections depending on positions and
momenta of other particles in the system.



Basically how do you get
around that and still retain SR?

First, I do not retain SR. SR means universal linear (tensor) Lorentz
transformations for all observables in all physical systems.
I do not retain that. SR means universal Minkowski background
space-time. I am not cool with that either.


Would you please specify a precise equation that doesn't
need 450 pages to explain, something like E=mc^2 that
we the people may examine. If you can't then deep-six
the book, don't waste time.

You can read just Chapter 1 (only twelve pages). The main point the
Lorentz transformations are derived from behavior of light pulses
(I haven't seen any other credible derivation yet). Then,
in the traditional approach,
these transformations are extended to all events in all systems
independent on their composition and interactions. Then this
huge generalization is cast into the form of universal Minkowski
space-time. Could you please fill this logical gap for me?
I bet you can't. The whole point of my book is that I am not
making this logical jump. I do not accept the idea of universality
of boost transformations. I derive these transformations
from dynamical equations, and find that they are not universal.



The points where I do agree with SR are
1) relativity postulate
2) Poincare group properties of inertial transformations.


You claimed not to understand that (GR) before....?

I did not mention GR in this post?



I said many times that in my approach instantaneous propagation of
interactions does not contradict causality (as in SR). There is no
"grandfather paradox" in my approach.


Eugene, becoming a theoretician is much more than gluing
a bunch of notes together. From chapter to chapter your
new book may overwhelm the reader with insignificant tactical
detail, but the strategy binding the purpose is lacking,
or merely implied. At the outset, you'll definitely need
to detail a concise strategy including a clear mathematical
definition of the new physicals law you propose and those
conventional laws you intend to revolutionize, ie an abstract.

Are you prepared to provide an abstract?
Ken

You can read PREFACE in my book. That's a good place to start.



Eugene

Ken S. Tucker wrote:



Eugene, becoming a theoretician is much more than gluing
a bunch of notes together. From chapter to chapter your
new book may overwhelm the reader with insignificant tactical
detail, but the strategy binding the purpose is lacking,
or merely implied. At the outset, you'll definitely need
to detail a concise strategy including a clear mathematical
definition of the new physicals law you propose and those
conventional laws you intend to revolutionize, ie an abstract.

Are you prepared to provide an abstract?
Ken

You can also read my papers published in refereed journals.
You can download these papers from www.geocities.com/meopemuk
They have short abstracts. If you like, we can discuss these
papers, especially

E. V. Stefanovich, "Is Minkowski
space-time compatible with
quantum mechanics?" Found. Phys. 32 (2002), 673.

which is a short version of the book.

Eugene.

Ken S. Tucker:
"FrediFizzx" wrote in message:

Photons do not have intrinsic invariant EM properties.
FrediFizzx

I argue they do. I think that because I underastand an
individual photon can be polarized, correct me if I'm
wrong please.

Sure thet do. In fact, the helicity is a lorentz invariant.
Furthermore, two polarization vectors which differ only by
a multiple of the momentum are the same photon.

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

[snip, most off topic, replace at will]

You haven't answered my question, from which I conclude that you don't
know any experiment which directly measures the speed of propagation
of interactions.

There is only one speed it's locally "c". This is proven
everyday in mundane ways.

You can read just Chapter 1 (only twelve pages). The main point the
Lorentz transformations are derived from behavior of light pulses
(I haven't seen any other credible derivation yet).

The Lorentz transform results from making absolute
velocity vanish,

U_i U^i =0 {i =1,2,3} U^i = dx^i/ds.

That's a simple equation, what value do you get?

Then,
in the traditional approach,
these transformations are extended to all events in all systems
independent on their composition and interactions. Then this
huge generalization is cast into the form of universal Minkowski
space-time. Could you please fill this logical gap for me?

I bet you can't.

See above.

The whole point of my book is that I am not
making this logical jump. I do not accept the idea of universality
of boost transformations. I derive these transformations
from dynamical equations, and find that they are not universal.



The points where I do agree with SR are
1) relativity postulate
2) Poincare group properties of inertial transformations.

You claimed not to understand that (GR) before....?

I did not mention GR in this post?

Then your relativity can only apply to non accelerating
particles. When you change your FoR to an accelerating
particle GR is implied.

Eugene