Why are the 'Fixed Stars' so FIXED?

On 25 Sep, 22:36, HW@....(Dr. Henri Wilson) wrote:
On Tue, 25 Sep 2007 05:47:38 -0700, George Dishman wrote:
On 23 Sep, 21:15, HW@....(Dr. Henri Wilson) wrote:
On Wed, 19 Sep 2007 22:16:11 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...

...but I cannot see how a gamma paticle can end up similarly dispersed....

Photons hit one atom on a detector but never that
next to it. They usually behave like bursts of
wavefronts in optical systems and zero-size
particles in detectors.

That is not an acceptible theory.

It is not a theory of any kind Henry. A theory
consists of a set of eqations that make
quanititative predictions of observed measurable
values.

What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

George, admit it. Nobody including you has much of a clue about the true nature
of light.

You obviously don't, QED is a complete theory
of light and gives arguably the most accurate
predictions of any physical theory.

Well you can't produce an interference pattern that way.

But they do Henry, a lot of astronomy is done that
way now. Take your EM program, think of it like
a string of umberellas and if one red umberella
wavefront passing through one telescope arrives at
the same time as another later red umberella
wavefront via the other telescope then you get
constructive interference. You have drawn the
pattern along a 1D 'ray' and the wavefronts are
normal to that but imagine what a 3D view would
look like. Who's to say how wide it could be?
The evidence is they cover the whole sphere at
emission.

Well I say the conclusions are a complete misinterpretation of the facts.

What I have written above are the facts, it is
up to you to interpret them. Photons only ever
cause a single detection in a sensor (perhaps
multiple electrons if avalanche technology is
used) and individual photons are subject to
all the normalrules of interference and
diffraction.

To get interference from two detectors spaced 100 metres apart requires
coherence. If a single photon stretches right across then what you said above
is wrong. A 100 m wide photon will strike more than one atom.
Even YOU should see that.

Of course. A simple telescope mirror has to
have the whole surface accurate to a fraction
of a wavelength because it all affects each
photon, but the photon only ever interacts
with _one_ atom when it finally hits the CCD
or photoelectric plate in a PM tube. You claim
to be familiar with the photoelectric effect
so you should know that already. The same is
true in the VLTI configuration, each photon
is affected by the paths through both
telescopes and all of both their surfaces, but
the detector still sees them as individual
particles. Those are the facts, what you have
to do is invent a theory (a set of equations)
that predicts that and the resulting interference
pattern.

You also said once before that the interference is cause byinteraction
between photons emitted from the two sides of the star.

No Henry, you said that. I said that was impossible
because they are uncorrelated. I think you spend
too much time inventing strawmen and forget what I
really said.

George, if the pattern is NOT caused by photons from both sides of the star,
how can it possibly privide information about radius change?

I explained that before. Photons(idividually) from
one side of the star create a pattern of intensity.
Photons from the other also create their own pattern,
again behaving individually, but the patterns are
displaced by a small amount, essentially the size
of the "image" formed by the telescope. Because one
is shifted relative to the other, the nulls don't
occur in the same place which affects the contrast
ratio of the fringe pattern. The actual analysis is
much more complex, involving limb darkening and lots
of other stuff but those are the basics.

It sounds as though all you are seeing is a fuzzy image of the star.

There are different ways to configure the detection.
In 2D mode what you get is an image crossed by
interference fringes. In 1D mode you get a graph
which looks like a static version of your EM
illustration, a windowed sine wave.

No Henry, you get interference when a single photon
is received at both ends of the antenna and combined.
If the path length difference is a multiple of the
wavelength you get constructive interference as usual.
That applies to each photon individually. The path
length difference for the phtotn includes a term that
depends on which side it came from since that slightly
changes the angle of the wavefront since it is
perpendicular to the line of site.

George, no matter how you try to wriggle out, a single photon from a star
cannot tell you anything about the star's radius.

Of course not, as I told you before, it is the
superposition of the displaced interference
patterns that carries the information....

I could just as easily carry information about varying light speed.

Don';t change the subject, it is the pattern that
provides the information as I ahve told you before,
not a single photon which is your strawman.

I don't accept that cepheids are really huff puff stars.

Tough, the radius is observed to vary and the
interferometric, photometric and integrated
velocity measurements are all similar.

.they are willusions.

Not according to ballistic theory, it says
they are valid.

I hope you are impressed George.

No, ballistic theory is wrong as shown by Sagnac
but I am pleased that you have learnt a little
local astrophysics, that's what this game is all
about ;-)

Sagnac disproves SR. ...

Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

I know more about calculus than you do.

ROFL, you just tried to tell me equation [3]
should be (c+v)/n, obviously you cannot even
differentiate an exponential :-)

I prefer to avoid it where possible.

Hardly surprising given your incompetence, but
then that's physics anyway and all you want to
do is talk philosophy.

George

On Wed, 26 Sep 2007 01:28:34 -0700, George Dishman
wrote:

On 25 Sep, 22:36, HW@....(Dr. Henri Wilson) wrote:
On Tue, 25 Sep 2007 05:47:38 -0700, George Dishman wrote:

What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

George, admit it. Nobody including you has much of a clue about the true nature
of light.

You obviously don't, QED is a complete theory
of light and gives arguably the most accurate
predictions of any physical theory.

George, QED doesn't tell us anything about the nature of light in transit.
The only facts we know about light relate to its arrival.
Light spends almost all of its time traveling.


To get interference from two detectors spaced 100 metres apart requires
coherence. If a single photon stretches right across then what you said above
is wrong. A 100 m wide photon will strike more than one atom.
Even YOU should see that.

Of course. A simple telescope mirror has to
have the whole surface accurate to a fraction
of a wavelength because it all affects each
photon, but the photon only ever interacts
with _one_ atom when it finally hits the CCD
or photoelectric plate in a PM tube.

You're the one becoming incoherent now George.

You claim
to be familiar with the photoelectric effect
so you should know that already. The same is
true in the VLTI configuration, each photon
is affected by the paths through both
telescopes and all of both their surfaces, but
the detector still sees them as individual
particles. Those are the facts, what you have
to do is invent a theory (a set of equations)
that predicts that and the resulting interference
pattern.

You aren't talking about interference. You're talking about a simple image.


much more complex, involving limb darkening and lots
of other stuff but those are the basics.

It sounds as though all you are seeing is a fuzzy image of the star.

There are different ways to configure the detection.
In 2D mode what you get is an image crossed by
interference fringes. In 1D mode you get a graph
which looks like a static version of your EM
illustration, a windowed sine wave.

You've lost me George.


Of course not, as I told you before, it is the
superposition of the displaced interference
patterns that carries the information....

I could just as easily carry information about varying light speed.

Don';t change the subject, it is the pattern that
provides the information as I ahve told you before,
not a single photon which is your strawman.

Here we go again...every time I ask George a difficult question he accuses me
of changing the subject...


Tough, the radius is observed to vary and the
interferometric, photometric and integrated
velocity measurements are all similar.

.they are willusions.

Not according to ballistic theory, it says
they are valid.

bull


I hope you are impressed George.

No, ballistic theory is wrong as shown by Sagnac
but I am pleased that you have learnt a little
local astrophysics, that's what this game is all
about ;-)

Sagnac disproves SR. ...

Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

I know more about calculus than you do.

ROFL, you just tried to tell me equation [3]
should be (c+v)/n, obviously you cannot even
differentiate an exponential :-)

What I suggested has nothing to do wtih calculus.
George, you're problem is you copy maths from websites to try to appear
knowledgable when you really haven't a clue what it all means.

I prefer to avoid it where possible.

Hardly surprising given your incompetence, but
then that's physics anyway and all you want to
do is talk philosophy.

George, I suggest you take a real course in physics instead of dreaming you
already have.

George

Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm

On Wed, 26 Sep 2007 01:11:39 -0700, George Dishman
wrote:

On 25 Sep, 22:53, HW@....(Dr. Henri Wilson) wrote:


Of course, the corona for example.

and that layer emits plenty of light.

Oops, no. You forgot Kirchofff's Law again, if it is
really transparent it cannot emit at all.

George, absorption is exponential.
If the layer absorbs only 10% per 100 kms, about 36% will still pass through
1000 kms. The layer will have an emissivity 0 and will still radiate.

See the next paragraph:

In practice
both the outer layer just above the photosphere and
the corona show spectral lines. The upper surface is
cooler than the photosphere so they appear in
absorption while the corona is hotter and the lines
show as emission. What we have been discussing re
temperature is the background continuum excluding the
lines.

There is no continuum emission from the 'transparent'
layer but where there is a resonance and the opacity
is higher at that specific frequency, we see absorption.

That's a funny paragraph.
'opacity' ...'specific frequency' ...I didn't know these had the same physical
dimensions...

...yes..and I have been pointing out that ADoppler can cause a considerable
shift in the planck curve.

And I have been pointing that while it _could_, we
know it _doesn't_, the actual shift is only 0.01%
and such a small shift doesn't affect temperature
determination.

If the shift is only 0.01%, so what?

Note also that if the frequency shift were larger,
it would no longer be a black body curve but in
practice the shift is so small that is negligible.

Many stars do not exhibit black body curves so how would you now if they were
shifted or not.

George

Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm

On 26 Sep, 11:44, HW@....(clueless Henri Wilson) wrote:
On Wed, 26 Sep 2007 01:28:34 -0700, George Dishman wrote:
On 25 Sep, 22:36, HW@....(Dr. Henri Wilson) wrote:
On Tue, 25 Sep 2007 05:47:38 -0700, George Dishman wrote:
What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

George, admit it. Nobody including you has much of a clue about the true nature
of light.

You obviously don't, QED is a complete theory
of light and gives arguably the most accurate
predictions of any physical theory.

George, QED doesn't tell us anything about the nature of light in transit.

You are back to the old philosophical debate
about whether something exists when it is not
being observed. QED tells you all the _science_.

The only facts we know about light relate to its arrival.
Light spends almost all of its time traveling.

To get interference from two detectors spaced 100 metres apart requires
coherence. If a single photon stretches right across then what you said above
is wrong. A 100 m wide photon will strike more than one atom.
Even YOU should see that.

Of course. A simple telescope mirror has to
have the whole surface accurate to a fraction
of a wavelength because it all affects each
photon, but the photon only ever interacts
with _one_ atom when it finally hits the CCD
or photoelectric plate in a PM tube.

You're the one becoming incoherent now George.

I clearly overestimated your knowledge again, I
thought you knew something about the photoelectric
effect. These are simple statements of fact.

You claim
to be familiar with the photoelectric effect
so you should know that already. The same is
true in the VLTI configuration, each photon
is affected by the paths through both
telescopes and all of both their surfaces, but
the detector still sees them as individual
particles. Those are the facts, what you have
to do is invent a theory (a set of equations)
that predicts that and the resulting interference
pattern.

You aren't talking about interference.

Yes I am, interferometers are regularly used
on astronomical telescopes.

You're talking about a simple image.

An image is formed by interference Henry, Paul
told you how.

much more complex, involving limb darkening and lots
of other stuff but those are the basics.

It sounds as though all you are seeing is a fuzzy image of the star.

There are different ways to configure the detection.
In 2D mode what you get is an image crossed by
interference fringes. In 1D mode you get a graph
which looks like a static version of your EM
illustration, a windowed sine wave.

You've lost me George.

I warned you it was complex, I'll try to
upload some synthesised images so you can
see what I mean but it won't be for a few
days.

Of course not, as I told you before, it is the
superposition of the displaced interference
patterns that carries the information....

I could just as easily carry information about varying light speed.

Don';t change the subject, it is the pattern that
provides the information as I ahve told you before,
not a single photon which is your strawman.

Here we go again...every time I ask George a difficult question he accuses me
of changing the subject...

You didn't ask a question, you made a statement
about a different topic. What other things light
might tell us in addition to what we are discussing
is hardly useful.

Tough, the radius is observed to vary and the
interferometric, photometric and integrated
velocity measurements are all similar.

.they are willusions.

Not according to ballistic theory, it says
they are valid.

bull

A statement of fact Henry.

I hope you are impressed George.

No, ballistic theory is wrong as shown by Sagnac
but I am pleased that you have learnt a little
local astrophysics, that's what this game is all
about ;-)

Sagnac disproves SR. ...

Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

I know more about calculus than you do.

ROFL, you just tried to tell me equation [3]
should be (c+v)/n, obviously you cannot even
differentiate an exponential :-)

What I suggested has nothing to do wtih calculus.

So you don't even recognise a first order
differential equation when you see it!

George, you're problem is you copy maths from websites to try to appear
knowledgable when you really haven't a clue what it all means.

Henry, your problem is that you haven't learned
any maths, equation [3] is trivial schoolboy
stuff and I was writing ones like it when I was
13 years old. I still use equations like that in
my work regularly. You seem to think it is some
fancy PhD stuff but the truth is you will find
it in any decent secondary school textbook.
Looking at some web pages suggests The Aussie
level for this stuff should be around year 8
in secondary school.

For equation [3] we want the speed to change as
the light travels, We could write an equation so
for dv/ds or dv/dt, I choose dv/ds for a couple
of reasons, notably because the mechanism must
involve interactions which can be defined as a
number per distance based on a density.

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

where A and B are positive constants. In case
you don't recognise it, the bit on the left is a
derivative meaning the rate at which v changes
as a function of s, the distance travelled, and
derivatives are one aspect of caclulus.

We know the eventual speed is c/n and at that
speed the rate of change is zero so if the bit
in brackets is to be zero constant when v=c/n,
the value of B must be c/n so:

dv/ds = A * (c/n - v)

The left hand side has units of speed/distance
and (c/n - v) has units of speed so constant A
must have units of inverse distance. It is easier
then to define R = 1 / A and hence:

dv/ds = (c/n - v) / R

where now R has units of distance. Hey presto
Henry, you have equation [3]. Easy, wasn't it.

I prefer to avoid it where possible.

Hardly surprising given your incompetence, but
then that's physics anyway and all you want to
do is talk philosophy.

George, I suggest you take a real course in physics instead of dreaming you
already have.

I have an honours degree in physics already, and
it's a real one, not like your fake paper. What
_you_ need to do is take the Aussie equivalent of
an English "A Level" in maths, or at least get up
to your 8th grade level so you can understand
equations like the one above without me having to
explain them to you. With a bit of practice you
should be able to write them for yourself.

George

On 26 Sep, 11:49, HW@....(Dr. Henri Wilson) wrote:
On Wed, 26 Sep 2007 01:11:39 -0700, George Dishman wrote:
On 25 Sep, 22:53, HW@....(Dr. Henri Wilson) wrote:

Of course, the corona for example.

and that layer emits plenty of light.

Oops, no. You forgot Kirchofff's Law again, if it is
really transparent it cannot emit at all.

George, absorption is exponential.
If the layer absorbs only 10% per 100 kms, about 36% will still pass through
1000 kms. The layer will have an emissivity 0 and will still radiate.

See the next paragraph:

In practice
both the outer layer just above the photosphere and
the corona show spectral lines. The upper surface is
cooler than the photosphere so they appear in
absorption while the corona is hotter and the lines
show as emission. What we have been discussing re
temperature is the background continuum excluding the
lines.

There is no continuum emission from the 'transparent'
layer but where there is a resonance and the opacity
is higher at that specific frequency, we see absorption.

That's a funny paragraph.
'opacity' ...'specific frequency' ...I didn't know these had the same physical
dimensions...

Why would you imagine they would? The opacity
of the gas just above the photosphere is higher
at the specific frequency we call hydrogen Alpha
than at other frequencies between such resonances.
You even claimed you had taken pictures in H alpha.

...yes..and I have been pointing out that ADoppler can cause a considerable
shift in the planck curve.

And I have been pointing that while it _could_, we
know it _doesn't_, the actual shift is only 0.01%
and such a small shift doesn't affect temperature
determination.

If the shift is only 0.01%, so what?

So the amount of energy that falls outside a
bandpass filter from 2000 to 2400 nm because
of the shift is negligible hence the temperature
measurement based on that value is valid.

Note also that if the frequency shift were larger,
it would no longer be a black body curve but in
practice the shift is so small that is negligible.

Many stars do not exhibit black body curves so how would you now if they were
shifted or not.

The shift is measured from specific lines,
you cannot measure a shift of 0.01% in the
continuum which is never exactly a black
body because of the finite thickness. This
is all obvious stuff Henry.

George

On Wed, 26 Sep 2007 05:24:38 -0700, George Dishman
wrote:

On 26 Sep, 11:44, HW@....(clueless Henri Wilson) wrote:
On Wed, 26 Sep 2007 01:28:34 -0700, George Dishman wrote:
On 25 Sep, 22:36, HW@....(Dr. Henri Wilson) wrote:
On Tue, 25 Sep 2007 05:47:38 -0700, George Dishman wrote:
What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

George, admit it. Nobody including you has much of a clue about the true nature
of light.

You obviously don't, QED is a complete theory
of light and gives arguably the most accurate
predictions of any physical theory.

George, QED doesn't tell us anything about the nature of light in transit.

You are back to the old philosophical debate
about whether something exists when it is not
being observed. QED tells you all the _science_.

Once again you claim that modern physics can answer all the questions...
It cannot....
QED tells us nothing about light in transit.

The only facts we know about light relate to its arrival.
Light spends almost all of its time traveling.

To get interference from two detectors spaced 100 metres apart requires
coherence. If a single photon stretches right across then what you said above
is wrong. A 100 m wide photon will strike more than one atom.
Even YOU should see that.

Of course. A simple telescope mirror has to
have the whole surface accurate to a fraction
of a wavelength because it all affects each
photon, but the photon only ever interacts
with _one_ atom when it finally hits the CCD
or photoelectric plate in a PM tube.

You're the one becoming incoherent now George.

I clearly overestimated your knowledge again, I
thought you knew something about the photoelectric
effect. These are simple statements of fact.

You are continually contradicting yourself.
Ii cannot follow what you are saying any more. One minute we have a 100m wide
photon that is refelcted from both telescopes onto a screen to form an image,
now we have this huge photon striking only one atom.
You aren't making sense George.

You claim
to be familiar with the photoelectric effect
so you should know that already. The same is
true in the VLTI configuration, each photon
is affected by the paths through both
telescopes and all of both their surfaces, but
the detector still sees them as individual
particles. Those are the facts, what you have
to do is invent a theory (a set of equations)
that predicts that and the resulting interference
pattern.

You aren't talking about interference.

Yes I am, interferometers are regularly used
on astronomical telescopes.

You're talking about a simple image.

An image is formed by interference Henry, Paul
told you how.

Paul knows even lees than you do...

much more complex, involving limb darkening and lots
of other stuff but those are the basics.



I could just as easily carry information about varying light speed.

Don';t change the subject, it is the pattern that
provides the information as I ahve told you before,
not a single photon which is your strawman.

Here we go again...every time I ask George a difficult question he accuses me
of changing the subject...

You didn't ask a question, you made a statement
about a different topic. What other things light
might tell us in addition to what we are discussing
is hardly useful.

George, I am trying to explain to you that what it being interpreted as a
radius change is possibly just a light velocity change.


Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

I know more about calculus than you do.

ROFL, you just tried to tell me equation [3]
should be (c+v)/n, obviously you cannot even
differentiate an exponential :-)

What I suggested has nothing to do wtih calculus.

So you don't even recognise a first order
differential equation when you see it!

Of course I can......that has nothing to do with the velocity being c or c+v in
the 'medium'.

.....but your equation is worthless unless you derive functions for both n and
r....and it will be a horrible thing to solve....Much better to let a computer
solve it.

George, you're problem is you copy maths from websites to try to appear
knowledgable when you really haven't a clue what it all means.

Henry, your problem is that you haven't learned
any maths, equation [3] is trivial schoolboy
stuff and I was writing ones like it when I was
13 years old. I still use equations like that in
my work regularly. You seem to think it is some
fancy PhD stuff but the truth is you will find
it in any decent secondary school textbook.
Looking at some web pages suggests The Aussie
level for this stuff should be around year 8
in secondary school.

George, stop it.....you're making me laugh....

For equation [3] we want the speed to change as
the light travels, We could write an equation so
for dv/ds or dv/dt, I choose dv/ds for a couple
of reasons, notably because the mechanism must
involve interactions which can be defined as a
number per distance based on a density.

George, your equation is trivial.
What you have to do is derive functions for both n and r...... versus distance.
Then you can solve the differential equation.

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

...and dv/dt will also be = A * (B - v)
since vc.

where A and B are positive constants. In case
you don't recognise it, the bit on the left is a
derivative meaning the rate at which v changes
as a function of s, the distance travelled, and
derivatives are one aspect of caclulus.



We know the eventual speed is c/n and at that
speed the rate of change is zero so if the bit
in brackets is to be zero constant when v=c/n,
the value of B must be c/n so:

dv/ds = A * (c/n - v)

The left hand side has units of speed/distance
and (c/n - v) has units of speed so constant A
must have units of inverse distance. It is easier
then to define R = 1 / A and hence:

dv/ds = (c/n - v) / R

where now R has units of distance. Hey presto
Henry, you have equation [3]. Easy, wasn't it.



I prefer to avoid it where possible.

Hardly surprising given your incompetence, but
then that's physics anyway and all you want to
do is talk philosophy.

George, I suggest you take a real course in physics instead of dreaming you
already have.

I have an honours degree in physics already, and
it's a real one, not like your fake paper. What
_you_ need to do is take the Aussie equivalent of
an English "A Level" in maths, or at least get up
to your 8th grade level so you can understand
equations like the one above without me having to
explain them to you. With a bit of practice you
should be able to write them for yourself.

George, stop raving and derive n(s) and r(s) for the various situations please.

George

Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm

On Wed, 26 Sep 2007 05:32:46 -0700, George Dishman
wrote:

On 26 Sep, 11:49, HW@....(Dr. Henri Wilson) wrote:
On Wed, 26 Sep 2007 01:11:39 -0700, George Dishman wrote:
On 25 Sep, 22:53, HW@....(Dr. Henri Wilson) wrote:

Of course, the corona for example.

and that layer emits plenty of light.

Oops, no. You forgot Kirchofff's Law again, if it is
really transparent it cannot emit at all.

George, absorption is exponential.
If the layer absorbs only 10% per 100 kms, about 36% will still pass through
1000 kms. The layer will have an emissivity 0 and will still radiate.

See the next paragraph:

In practice
both the outer layer just above the photosphere and
the corona show spectral lines. The upper surface is
cooler than the photosphere so they appear in
absorption while the corona is hotter and the lines
show as emission. What we have been discussing re
temperature is the background continuum excluding the
lines.

There is no continuum emission from the 'transparent'
layer but where there is a resonance and the opacity
is higher at that specific frequency, we see absorption.

That's a funny paragraph.
'opacity' ...'specific frequency' ...I didn't know these had the same physical
dimensions...

Why would you imagine they would? The opacity
of the gas just above the photosphere is higher
at the specific frequency we call hydrogen Alpha
than at other frequencies between such resonances.
You even claimed you had taken pictures in H alpha.

Sorry I misread your statement...
It's OK..

...yes..and I have been pointing out that ADoppler can cause a considerable
shift in the planck curve.

And I have been pointing that while it _could_, we
know it _doesn't_, the actual shift is only 0.01%
and such a small shift doesn't affect temperature
determination.

If the shift is only 0.01%, so what?

So the amount of energy that falls outside a
bandpass filter from 2000 to 2400 nm because
of the shift is negligible hence the temperature
measurement based on that value is valid.

Note also that if the frequency shift were larger,
it would no longer be a black body curve but in
practice the shift is so small that is negligible.

Many stars do not exhibit black body curves so how would you now if they were
shifted or not.

The shift is measured from specific lines,
you cannot measure a shift of 0.01% in the
continuum which is never exactly a black
body because of the finite thickness. This
is all obvious stuff Henry.

George, let's get this straight.

You say temperature is estimated by comparing the energy arriving in the two
fliter bands and assuming a black body curve.

I say the average light in the two bands can come from slightly different
layers with different radial velocities. ADoppler can shift the light by much
more than 0.01% and might easily lead to false ratios in the two bands.

George

Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm

On 26 Sep, 23:09, HW@....(Dr. Henri Wilson) wrote:
On Wed, 26 Sep 2007 05:24:38 -0700, George Dishman wrote:
On 26 Sep, 11:44, HW@....(clueless Henri Wilson) wrote:
On Wed, 26 Sep 2007 01:28:34 -0700, George Dishman wrote:
On 25 Sep, 22:36, HW@....(Dr. Henri Wilson) wrote:
On Tue, 25 Sep 2007 05:47:38 -0700, George Dishman wrote:
What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

George, admit it. Nobody including you has much of a clue about the true nature
of light.

You obviously don't, QED is a complete theory
of light and gives arguably the most accurate
predictions of any physical theory.

George, QED doesn't tell us anything about the nature of light in transit.

You are back to the old philosophical debate
about whether something exists when it is not
being observed. QED tells you all the _science_.

Once again you claim that modern physics can answer all the questions...

No, it answers only the scientific questions, not
philosophical ones.

It cannot....
QED tells us nothing about light in transit.

QED addresses the relationship between emission
and detection. Whether the photon exists while
it is not being observed, i.e. in transit, is
a philosophical question. If the photon interacts
while in transit, that interaction is correctly
explained by QED.

To get interference from two detectors spaced 100 metres apart requires
coherence. If a single photon stretches right across then what you said above
is wrong. A 100 m wide photon will strike more than one atom.
Even YOU should see that.

Of course. A simple telescope mirror has to
have the whole surface accurate to a fraction
of a wavelength because it all affects each
photon, but the photon only ever interacts
with _one_ atom when it finally hits the CCD
or photoelectric plate in a PM tube.

You're the one becoming incoherent now George.

I clearly overestimated your knowledge again, I
thought you knew something about the photoelectric
effect. These are simple statements of fact.

You are continually contradicting yourself.
Ii cannot follow what you are saying any more. One minute we have a 100m wide
photon that is refelcted from both telescopes onto a screen to form an image,
now we have this huge photon striking only one atom.
You aren't making sense George.

Reality may not make sense to you Henry, what
you have just said is a pretty good summary of
what is observed. The VLTI system is a good
example but there are others where distributed
telescopes are linked to make interferometers
and they can work with single photons. The point
you make about the detection is what you yourself
were citing to Sean as the evidence that light is
in the form of particles, something we agee on
entirely, but Young's Slits, gratings and large
telescopes all show interference on individual
photons. Any attempt you make to produce a
particle based theory will need to address that.

much more complex, involving limb darkening and lots
of other stuff but those are the basics.

I could just as easily carry information about varying light speed.

Don';t change the subject, it is the pattern that
provides the information as I ahve told you before,
not a single photon which is your strawman.

Here we go again...every time I ask George a difficult question he accuses me
of changing the subject...

You didn't ask a question, you made a statement
about a different topic. What other things light
might tell us in addition to what we are discussing
is hardly useful.

George, I am trying to explain to you that what it being interpreted as a
radius change is possibly just a light velocity change.

And I have explained that the phase difference
through the two light paths for a single photon
can only depend on the speed over the last few
wavelengths prior to the surface of the telescopes.
Any speed variations prior to that are common to
both paths since we are talking about just one
photon. There is no point in repeating what you
said when I have already told you why it does not
have any effect.

Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

I know more about calculus than you do.

ROFL, you just tried to tell me equation [3]
should be (c+v)/n, obviously you cannot even
differentiate an exponential :-)

What I suggested has nothing to do wtih calculus.

So you don't even recognise a first order
differential equation when you see it!

Of course I can......that has nothing to do with the velocity being c or c+v in
the 'medium'.

No, it is the equation for "speed equalisation"
after its emission at c+v.

....but your equation is worthless unless you derive functions for both n and
r....and it will be a horrible thing to solve....Much better to let a computer
solve it.

Go ahead then, if you think the equation isn't
valid if I have to look up the refractive of
glass and instead it must be derived from first
principles, go ahead and write your program to
do it. Any conventional derivation would apply
QED of course but ballistic theory replaces that
so in your case you would have to write your own
derivation using equations [2] and [3].

George, you're problem is you copy maths from websites to try to appear
knowledgable when you really haven't a clue what it all means.

Henry, your problem is that you haven't learned
any maths, equation [3] is trivial schoolboy
stuff and I was writing ones like it when I was
13 years old. I still use equations like that in
my work regularly. You seem to think it is some
fancy PhD stuff but the truth is you will find
it in any decent secondary school textbook.
Looking at some web pages suggests The Aussie
level for this stuff should be around year 8
in secondary school.

George, stop it.....you're making me laugh....

Then stop making mathematical mistakes that
any competent 14 year old could correct.

For equation [3] we want the speed to change as
the light travels, We could write an equation so
for dv/ds or dv/dt, I choose dv/ds for a couple
of reasons, notably because the mechanism must
involve interactions which can be defined as a
number per distance based on a density.

George, your equation is trivial.
What you have to do is derive functions for both n and r...... versus distance.
Then you can solve the differential equation.

I don't need to derive the values at all, they
relate to actual situations so for example for
the pulsars we looked at, the value of n(s) is
measured using the frequency dependence. In
reality, r(s) cannot be measured since it
doesn't exist.

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

..and dv/dt will also be = A * (B - v)
since vc.

Yes, and it would be a good excercise for you to
see if you can work ou why I chose to use dv/ds
instead of dv/dt. You'll need to learn some
calculus to do that.

where A and B are positive constants. In case
you don't recognise it, the bit on the left is a
derivative meaning the rate at which v changes
as a function of s, the distance travelled, and
derivatives are one aspect of caclulus.



We know the eventual speed is c/n and at that
speed the rate of change is zero so if the bit
in brackets is to be zero constant when v=c/n,
the value of B must be c/n so:
dv/ds = A * (c/n - v)

The left hand side has units of speed/distance
and (c/n - v) has units of speed so constant A
must have units of inverse distance. It is easier
then to define R = 1 / A and hence:

dv/ds = (c/n - v) / R

where now R has units of distance. Hey presto
Henry, you have equation [3]. Easy, wasn't it.



Before smiling, bear in mind you have been
touting your handwaving for ten years or more
and in all that time you were never able to
produce that equation while it only took me
minutes.

I prefer to avoid it where possible.

Hardly surprising given your incompetence, but
then that's physics anyway and all you want to
do is talk philosophy.

George, I suggest you take a real course in physics instead of dreaming you
already have.

I have an honours degree in physics already, and
it's a real one, not like your fake paper. What
_you_ need to do is take the Aussie equivalent of
an English "A Level" in maths, or at least get up
to your 8th grade level so you can understand
equations like the one above without me having to
explain them to you. With a bit of practice you
should be able to write them for yourself.

George, stop raving and derive n(s) and r(s) for the various situations please.

I have already given you the link to the dispersion
equation for n(s) from the Jodrell Bank site. There
is no equation for r(s) but if you remember we found
an upper value for the mean of r(s) from the pulsar
data. This is supposed to be your theory so you have
to write it yourself Henry, but I can tell you how I
would approach it if I were you.

Work in the rest frame of the material and suppose we
want to calculate r(s) for the ISM (in a solid the
atoms are somewhat fixed so you need to consider
vibrational modes and such like). A photon moving at
initial speed c+Vi is propagating through space and
hits a charged particle of mass M which is at rest.
The photon carries momentum P which is transferred to
the particle causing it to move with speed Va. Note
that we need Va Vi for Vi 0. The particle then
emits a photon with speed c relative to the particle,
thus at speed Va+c which carries off some momentum P.
The recoil slows the particle to Vb hence after the
interaction some momentum has been transferred to the
particle and the photon which was initially moving
faster than c has lost momentum and has been slowed
which looks good.

To turn that into an equation derived from your theory,
you need to first provide an equation that relates the
photon momentum to its speed. Use that to find the
initial P value, say P=f(c+Vi), use Va = P/M for the
particle and then conservation of momentum gives
P = M*Vb + f(c+Va).

To get r(s) finally, you need to work out the number
of such interactions per unit distance as a function
of the density in a similar fashion to the pulsar
formula I showed you for the refractive index.

There is a problem unfortunately, if you start with
Vi 0 (e.g. a "c-v" photon) then the particle moves
more slowly and the recoil means the same process
speeds up the photon as you want, however the particle
still gets a positive speed and the emitted photon has
speed c after the interaction. Solving that one is
your task, this is supposed to be your theory after all
so I can't do it all for you.

George

On Thu, 27 Sep 2007 01:16:58 -0700, George Dishman
wrote:

On 26 Sep, 23:09, HW@....(Dr. Henri Wilson) wrote:
On Wed, 26 Sep 2007 05:24:38 -0700, George Dishman wrote:


You are back to the old philosophical debate
about whether something exists when it is not
being observed. QED tells you all the _science_.

Once again you claim that modern physics can answer all the questions...

No, it answers only the scientific questions, not
philosophical ones.

your version does.

It cannot....
QED tells us nothing about light in transit.

QED addresses the relationship between emission
and detection. Whether the photon exists while
it is not being observed, i.e. in transit, is
a philosophical question. If the photon interacts
while in transit, that interaction is correctly
explained by QED.

Oh crap.
Of course light exists is some form during its lifetime of travel.
That form can be described in terms of OUR physical 3D, 1T universe.
This is pure physics, not philosophy.

The Einstein claim that all starlight in the universe travels miraculously at
exactly c wrt litlte planet Earth is philosophy....the fairyland type...


You are continually contradicting yourself.
Ii cannot follow what you are saying any more. One minute we have a 100m wide
photon that is refelcted from both telescopes onto a screen to form an image,
now we have this huge photon striking only one atom.
You aren't making sense George.

Reality may not make sense to you Henry, what
you have just said is a pretty good summary of
what is observed. The VLTI system is a good
example but there are others where distributed
telescopes are linked to make interferometers
and they can work with single photons. The point
you make about the detection is what you yourself
were citing to Sean as the evidence that light is
in the form of particles, something we agee on
entirely, but Young's Slits, gratings and large
telescopes all show interference on individual
photons. Any attempt you make to produce a
particle based theory will need to address that.

OK, it appears that photons 'spread out' as they travel.
I have no problem with that except to add that is seems logical that they might
also lose a little energy in doing so and appear to be red shifted as a result.



You didn't ask a question, you made a statement
about a different topic. What other things light
might tell us in addition to what we are discussing
is hardly useful.

George, I am trying to explain to you that what it being interpreted as a
radius change is possibly just a light velocity change.

And I have explained that the phase difference
through the two light paths for a single photon
can only depend on the speed over the last few
wavelengths prior to the surface of the telescopes.
Any speed variations prior to that are common to
both paths since we are talking about just one
photon. There is no point in repeating what you
said when I have already told you why it does not
have any effect.

....and I have tried to explain to you that one photon cannot reveal anything
about any radius change.


What I suggested has nothing to do wtih calculus.

So you don't even recognise a first order
differential equation when you see it!

Of course I can......that has nothing to do with the velocity being c or c+v in
the 'medium'.

No, it is the equation for "speed equalisation"
after its emission at c+v.

....but your equation is worthless unless you derive functions for both n and
r....and it will be a horrible thing to solve....Much better to let a computer
solve it.

Go ahead then, if you think the equation isn't
valid if I have to look up the refractive of
glass and instead it must be derived from first
principles, go ahead and write your program to
do it. Any conventional derivation would apply
QED of course but ballistic theory replaces that
so in your case you would have to write your own
derivation using equations [2] and [3].

George, YOUR equation is trivial. The hard part is finding n(s) and r(s).

relate to actual situations so for example for
the pulsars we looked at, the value of n(s) is
measured using the frequency dependence. In
reality, r(s) cannot be measured since it
doesn't exist.

your opinion....

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

..and dv/dt will also be = A * (B - v)
since vc.

Yes, and it would be a good excercise for you to
see if you can work ou why I chose to use dv/ds
instead of dv/dt. You'll need to learn some
calculus to do that.



Obviously n(s) and r(s) are much easier to consider as functions of distance,
rather than time.

where now R has units of distance. Hey presto
Henry, you have equation [3]. Easy, wasn't it.



Before smiling, bear in mind you have been
touting your handwaving for ten years or more
and in all that time you were never able to
produce that equation while it only took me
minutes.

George, I have a mind that does these things automatically. I don't need any
equations until I get the model right.
You being an engineer require an equation BEFORE you get the model right.
Physicists provide engineers with equations.

George, stop raving and derive n(s) and r(s) for the various situations please.

I have already given you the link to the dispersion
equation for n(s) from the Jodrell Bank site. There
is no equation for r(s) but if you remember we found
an upper value for the mean of r(s) from the pulsar
data. This is supposed to be your theory so you have
to write it yourself Henry, but I can tell you how I
would approach it if I were you.

There is NO clearcut equation for either.
There are many different situations regarding the common EM sphere around
binary pairs.
I suppose it wouldn't be too hard to describe what YOU are likening to a
refractive index, n(s), as something like: 1-n = (1-n(o)).K/(s^3)....but that
would be an oversimplification.

r(s) would be hard to describe since it involves statistics. Light randomly
speeds up and slows down as it travels....
We can regrad the universe as being turbulent like any rare gas.
However, we can state that type(2) unification AFTER LEAVING THE STAR'S SPHERE
requires that c+v+u - c-v+u. in other words the v - zero with distance. After
unification, the velocity 'u' will also vary depending on the properties of the
particular region of space the light happens to be in.

Work in the rest frame of the material and suppose we
want to calculate r(s) for the ISM (in a solid the
atoms are somewhat fixed so you need to consider
vibrational modes and such like). A photon moving at
initial speed c+Vi is propagating through space and
hits a charged particle of mass M which is at rest.
The photon carries momentum P which is transferred to
the particle causing it to move with speed Va. Note
that we need Va Vi for Vi 0. The particle then
emits a photon with speed c relative to the particle,
thus at speed Va+c which carries off some momentum P.
The recoil slows the particle to Vb hence after the
interaction some momentum has been transferred to the
particle and the photon which was initially moving
faster than c has lost momentum and has been slowed
which looks good.

I think that approach is too conventional.
It can't account for the speeding up of light...even though it can explain the
cosmic redshift.

To turn that into an equation derived from your theory,
you need to first provide an equation that relates the
photon momentum to its speed. Use that to find the
initial P value, say P=f(c+Vi), use Va = P/M for the
particle and then conservation of momentum gives
P = M*Vb + f(c+Va).

To get r(s) finally, you need to work out the number
of such interactions per unit distance as a function
of the density in a similar fashion to the pulsar
formula I showed you for the refractive index.

This is still the 'refractive index' approach.
I maintain that unification involves MORE than ordinary matter.

If a photon is a discreet particle, it must interact with its surroundings in
order to change speed ...up or down...
I still like the idea that a photon is basically a rapidly rotating pair of
charges carving out a helix or some kind of spatial patern as they travel.
Polarization has to be explained though.

There is a problem unfortunately, if you start with
Vi 0 (e.g. a "c-v" photon) then the particle moves
more slowly and the recoil means the same process
speeds up the photon as you want, however the particle
still gets a positive speed and the emitted photon has
speed c after the interaction. Solving that one is
your task, this is supposed to be your theory after all
so I can't do it all for you.

At least you are now trying rather than rejecting it outright....
Thanks George...

George

Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm

On 27 Sep, 22:55, HW@....(Dr. Henri Wilson) wrote:
On Thu, 27 Sep 2007 01:16:58 -0700, George Dishman wrote:
On 26 Sep, 23:09, HW@....(Dr. Henri Wilson) wrote:
On Wed, 26 Sep 2007 05:24:38 -0700, George Dishman wrote:

You are back to the old philosophical debate
about whether something exists when it is not
being observed. QED tells you all the _science_.

Once again you claim that modern physics can answer all the questions...

No, it answers only the scientific questions, not
philosophical ones.

your version does.

Right.

It cannot....
QED tells us nothing about light in transit.

QED addresses the relationship between emission
and detection. Whether the photon exists while
it is not being observed, i.e. in transit, is
a philosophical question. If the photon interacts
while in transit, that interaction is correctly
explained by QED.

Oh crap.
Of course light exists is some form during its lifetime of travel.
That form can be described in terms of OUR physical 3D, 1T universe.
This is pure physics, not philosophy.

No, physics is about what can be measured so
by definition, what happens when it isn't
measured is not physics. What we can say is
that certain properties are conserved which
implies that they have the same values when
not being measured, but that is only an
implication, it cannot be confirmed obviously.

The Einstein claim that all starlight in the universe travels miraculously at
exactly c wrt litlte planet Earth is philosophy....the fairyland type...

Of course, your pathetic strawman is pure
fairyland stuff. In the real world, SR says
the speed is c in _all_inertial_ frames, and
Earth isn't inertial, while there are an
infinite number of inertial frames.

You are continually contradicting yourself.
Ii cannot follow what you are saying any more. One minute we have a 100m wide
photon that is refelcted from both telescopes onto a screen to form an image,
now we have this huge photon striking only one atom.
You aren't making sense George.

Reality may not make sense to you Henry, what
you have just said is a pretty good summary of
what is observed. The VLTI system is a good
example but there are others where distributed
telescopes are linked to make interferometers
and they can work with single photons. The point
you make about the detection is what you yourself
were citing to Sean as the evidence that light is
in the form of particles, something we agee on
entirely, but Young's Slits, gratings and large
telescopes all show interference on individual
photons. Any attempt you make to produce a
particle based theory will need to address that.

OK, it appears that photons 'spread out' as they travel.

Do a Young's Slits experiment. First put
the source 10cm from slits 1cm apart and
note you get an interference pattern. Then
put the source 10m from slits 1m apart and
note a similar pattern forms. It is the same
as before, just scaled, if you also change
the wavelength by a factor of 10. If you
could set up the slits 1 light year apart
and 10 light years from a source, you would
again get an interference pattern and VLTI
with "slits" 102m apart many light years
from the source works exactly the same way.

I have no problem with that except to add that is seems logical that they might
also lose a little energy in doing so and appear to be red shifted as a result.

The spreading out would be inverse square so
you would expect photons in the 10m experiment
to have 0.01% of the energy in the 10cm version,
but light is particulate, all the energy lands
at a single point in the detector.

You didn't ask a question, you made a statement
about a different topic. What other things light
might tell us in addition to what we are discussing
is hardly useful.

George, I am trying to explain to you that what it being interpreted as a
radius change is possibly just a light velocity change.

And I have explained that the phase difference
through the two light paths for a single photon
can only depend on the speed over the last few
wavelengths prior to the surface of the telescopes.
Any speed variations prior to that are common to
both paths since we are talking about just one
photon. There is no point in repeating what you
said when I have already told you why it does not
have any effect.

...and I have tried to explain to you that one photon cannot reveal anything
about any radius change.

And I have told you that it is the contrast ratio of
the overlapping interference patterns that is used
to determine the radius. Interference applies to
each photon individually so is unaffected by the
speed in space but it is the cumulative distribution
that gives us the information.

What I suggested has nothing to do wtih calculus.

So you don't even recognise a first order
differential equation when you see it!

Of course I can......that has nothing to do with the velocity being c or c+v in
the 'medium'.

No, it is the equation for "speed equalisation"
after its emission at c+v.

....but your equation is worthless unless you derive functions for both n and
r....and it will be a horrible thing to solve....Much better to let a computer
solve it.

Go ahead then, if you think the equation isn't
valid if I have to look up the refractive of
glass and instead it must be derived from first
principles, go ahead and write your program to
do it. Any conventional derivation would apply
QED of course but ballistic theory replaces that
so in your case you would have to write your own
derivation using equations [2] and [3].

George, YOUR equation is trivial. The hard part is finding n(s) and r(s).

My equation is the one given by ballistic theory.
For n(s) and r(s) you need to look at material
science as well. For example to predict the
refractive index of diamond, you need to
understand the crystal structure, and that isn't
the province of your theory.

relate to actual situations so for example for
the pulsars we looked at, the value of n(s) is
measured using the frequency dependence. In
reality, r(s) cannot be measured since it
doesn't exist.

your opinion....

And that of all scientists, but it means you
can't find the work done for you, you need to
work it out yourself.

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

..and dv/dt will also be = A * (B - v)
since vc.

Yes, and it would be a good excercise for you to
see if you can work ou why I chose to use dv/ds
instead of dv/dt. You'll need to learn some
calculus to do that.



Obviously n(s) and r(s) are much easier to consider as functions of distance,
rather than time.

Yes, that's one reason and perfectly valid, but
there is another that comes purely from calculus.
It's a bit more subtle and not a rigid requirement,
it just make life a bit easier.

where now R has units of distance. Hey presto
Henry, you have equation [3]. Easy, wasn't it.



Before smiling, bear in mind you have been
touting your handwaving for ten years or more
and in all that time you were never able to
produce that equation while it only took me
minutes.

George, I have a mind that does these things automatically. I don't need any
equations until I get the model right.

The equations _are_ the model Henry, no equations
means no theory.

You being an engineer require an equation BEFORE you get the model right.
Physicists provide engineers with equations.

Exactly! So if you want to pretend you are a
physicist, _you_ should be providing _me_ with
rthe equations, not the other way round as it
is at the moment.

George, stop raving and derive n(s) and r(s) for the various situations please.

I have already given you the link to the dispersion
equation for n(s) from the Jodrell Bank site. There
is no equation for r(s) but if you remember we found
an upper value for the mean of r(s) from the pulsar
data. This is supposed to be your theory so you have
to write it yourself Henry, but I can tell you how I
would approach it if I were you.

There is NO clearcut equation for either.

http://www.jb.man.ac.uk/~pulsar/tuto...ut/node14.html

There are many different situations regarding the common EM sphere around
binary pairs.

I am talking about the ISM, not the region near
the star.

I suppose it wouldn't be too hard to describe what YOU are likening to a
refractive index, n(s), as something like: 1-n = (1-n(o)).K/(s^3)....but that
would be an oversimplification.

The equation is given on the page above.

r(s) would be hard to describe since it involves statistics. Light randomly
speeds up and slows down as it travels....
We can regrad the universe as being turbulent like any rare gas.
However, we can state that type(2) unification AFTER LEAVING THE STAR'S SPHERE
requires that c+v+u - c-v+u. in other words the v - zero with distance. After
unification, the velocity 'u' will also vary depending on the properties of the
particular region of space the light happens to be in.

The same applies within the sphere but you cannot
assume r(s) is constant. An equation deriving r(s)
from density could give you the overall formula by
noting that the stellar wind is likely to fall off
roughly as the inverse square (not quite since the
material is being slowed by gravity but good for
a first approximation).

Work in the rest frame of the material and suppose we
want to calculate r(s) for the ISM (in a solid the
atoms are somewhat fixed so you need to consider
vibrational modes and such like). A photon moving at
initial speed c+Vi is propagating through space and
hits a charged particle of mass M which is at rest.
The photon carries momentum P which is transferred to
the particle causing it to move with speed Va. Note
that we need Va Vi for Vi 0. The particle then
emits a photon with speed c relative to the particle,
thus at speed Va+c which carries off some momentum P.
The recoil slows the particle to Vb hence after the
interaction some momentum has been transferred to the
particle and the photon which was initially moving
faster than c has lost momentum and has been slowed
which looks good.

I think that approach is too conventional.

It is actually more general than you think.

It can't account for the speeding up of light...even though it can explain the
cosmic redshift.

See later.

To turn that into an equation derived from your theory,
you need to first provide an equation that relates the
photon momentum to its speed. Use that to find the
initial P value, say P=f(c+Vi), use Va = P/M for the
particle and then conservation of momentum gives
P = M*Vb + f(c+Va).

To get r(s) finally, you need to work out the number
of such interactions per unit distance as a function
of the density in a similar fashion to the pulsar
formula I showed you for the refractive index.

This is still the 'refractive index' approach.

It is a particulate approach, nothing more.

I maintain that unification involves MORE than ordinary matter.

Typically the particles would be assumed to be
the species observed, mostly electrons, protons,
alpha particles etc. but there is no reason why
it shouldn't apply to particles other than
normal matter, all you need is the value of M
which is the coefficient of the momentum to
speed function for the particle.

If a photon is a discreet particle, it must interact with its surroundings in
order to change speed ...up or down...

Yep, and the above deals with any particulate
form of "surroundings".

I still like the idea that a photon is basically a rapidly rotating pair of
charges carving out a helix or some kind of spatial patern as they travel.
Polarization has to be explained though.

Polarisation is easy, your model is similar to a
conventional circularly polarised signal and two
can be combined to give linear polarisation. Where
you have a problem is conceptually since your are
trying to give a particle-based theory so there
are no fields, the photon _is_ the field, so you
end up with photons giving off photons, and
experimentally you have to explain why we cannot
detect the individual charges.

There is a problem unfortunately, if you start with
Vi 0 (e.g. a "c-v" photon) then the particle moves
more slowly and the recoil means the same process
speeds up the photon as you want, however the particle
still gets a positive speed and the emitted photon has
speed c after the interaction. Solving that one is
your task, this is supposed to be your theory after all
so I can't do it all for you.

At least you are now trying rather than rejecting it outright....
Thanks George...

Oh I reject it outright of course, Sagnac proves
the whole idea is wrong from the start, but I am
happy to teach you how to do physics for yourself.

So the question is can you do the physics and solve
the problem? I have given you a hint, if you use
the same approach of breaking the equation down
logically as I did to write equation [3], you
will find the solution. The key is that you apply
the method to work out the equation for the
momentum of a photon as a function of its speed.
See if you can do it yourself this time, if not
I'll tell you the answer and show you how to do
it again.

George