Why are the 'Fixed Stars' so FIXED?

On 10 May, 00:58, HW@....(Henri Wilson) wrote:
On 9 May 2007 05:52:35 -0700, George Dishman wrote:
On 9 May, 00:19, HW@....(Henri Wilson) wrote:
On 8 May 2007 01:39:46 -0700, George Dishman wrote:
On 7 May, 23:52, HW@....(Henri Wilson) wrote:
Maximum observed velocity is ~300km/s for contact binaries
or 0.1%c. That is also the catch-up ratio so the bunching is
asymptotic to reducing the spacing by 0.1% at most. Think
of your findings on the pulsars if you have trouble following
the logic.

That's only the VDoppler component.

No, it is TDoppler. How many times do I have to correct you
on that?

George, I know TDoppler is the overall cause.
When I said, "That's only the VDoppler component", I was implying that the
VDoppler component was greater than the ADoppler component, which can be
ignored.

OK, in that case we agree. Please try to say TDoppler if that's
what you mean, it just wastes a lot of time clearing this up
each time.

No it doesn't George. You are telling little fibs again.
The photons keep moving at c+v for a lot longer than the 'ends of each photon'.
It's all so simple really.

Nicely put, the beginning, end and middle of each photon
move at (c+v)/10000 while the mean speed of the photon
is (c+v).

No you've gotten it all wrong again George.

I think you meant c+(v/10000)....but it doesn't even do that for very long.

I meant (c+v)/10000 but c+(v/10000) is also possible, your
theory is self-contradictory which means if I assume c+v I
can use it to prove (c+v)/10000 or vice versa or maybe that
black is white. The trouble with self-contradictory theories
is that they produce results that violate their own postulates
so the number you get depends on what route you take.

See, George, you have been missing the point all along.

What are you talking about? What I said was correct.

The
whole photon settles down to a fixed length that is shorter than when it was
emitted by L'=Le(1-Ka), where a is the radial acceleration of the source at the
point of emission.

Henry, there is only _one_ equation for the speed in your
theory and it applies to _all_ parts so K=1.

No you've gotten it all wrong again George.

No Henry, you just don't understand how physical laws can be
used as tools so that one assumption, say c+v, leads to other
conclusions like the Doppler equation by purely mathematical
means.

What a strange this to say!

Not really, K=1 can be derived purely mathematically and you
are ignoring that which produces your inconsistency.

This is just the procedure I'm following in relation to star brightness curves
and the BaTh.

No, you are plucking a value of K out of thin air instead of
deriving it using Fourier analysis (or some other equivalent
approach, there's more than one way).

You claimed elsewhere you knew how to use a Fourier
transform (which I doubt but never mind) so just apply
it to a pulse modulated carrier and see what you get
if you apply your Doppler equation to the components.
Reverse transform the frequency shifted elements to
get the received waveform as usual.

An individual photon has intrinsic properties that are not part of the group
bunching process. Hiwever it is still subject to ADoppler, in a small way.

It's all so simple if you open up your mind George.

Of course I can believe in anything if I allow for fantasies
but raw maths rules out your handwaving crap and this is
a science group, not sci-fi.

You like to model the maths to suit yourself.

You don't "model the maths" Henry, you use maths to
model the real world.

My theory is perfectly mathematically sound.

No, it is mathematically self-contradictory, I have
explained why dozens of times. Do a Fourier transform
and see for yourself.

An EM FoR is ...

a mathematical coordinate system with no physical existence
being used to defines locations and time of EM phenomena.

It ''''loosely''''' defines EM speed in that FoR

No, I can describe the speed of light in my office using a
coordinate system centred on the barycentre of the Bullet
Cluster, but the cluster does not define the speed in any
way whatsoever.

OK 'defines' wasn't the best word.

"Describes" or "labels" would be better but the key
point is that the FoR has no physical existence.

In an EM FoR, light speed will TEND TOWARDS c/n wrt the frame's 'EM centre'.

c/n wrt the material which produces the refractive index n
as measured using a frame of reference in which that
material is at rest would probably be the clearest way to
state that.

The latter is another Wilsonian pseudo-geometric term that describes a kind of
average influence exerted by all the 'substance' inside the frame (ie., matter
and fields) on all light originating in or passing through it.

I trust that is now settled.

Just say 'matter' when you mean it and the problem goes away.

For other angles the equation is N(lambda= D[sin(theta)/(c+u)-sin(phi)/(c+v)]

Yes, I was assuming the first order result in my other
replies too. In general the BaTh grating equation is:

N * lambda_r = D * sin(phi)

Why do you want to use lamba_r?

Henry, do you understand what it means to put a variable
on the left hand side of an equation? Perhaps I should have
written it as:

lambda_r = D * sin(phi) / N

but I kept it similar to yours to help you follow. I am only
"using" D and phi, both of which I can measure. lambda_r is
the result.

You don't know what its value is unless you
know the reflected light speed exactly. ..but you know lambda_i because it's is
absolute and universal.

Wrong, measuring D, the grating spacing, and phi, the
deflection angle, tells me lambda_r. I know nothing more
than that. You certainly don't know lambda_i because that
depends on the source, gravitational redshift, cosmogical
redshift, speed equalisation, material conditions and
magnetic fields in the source and so on.

George, you obviously didn't follow my diagram. It describes BaTh not aether
theory. Have another look...

I didn't mention an aether. I'm fairly happy with your diagram.

The criterion is TIME not distance. My 'lambda' is your Lambda_i.
The time taken for ray 2 reach the grating after the previous wavecrest (front)
from ray 1 has been reflected to the end of line 'x' is Lambda_i/(c+v).
The time for ray 1's reflection to travel distance 'x' is Dsin(phi)/(c+u)

...these two times are equal.

The time for ray 1 is the distance from ruling 1 to the point on
the screen divided by the speed of ray 1. The time for ray 2 is
the distance from ruling 2 to the same point on the screen
divided by the speed of ray 2. You are assuming that the speed
of the two rays is the same, i.e. u1 = u2. That may not be the
case if the deflection angles differ as might be the case if the
screen were very close to the grating, however, if we assume
the rays are close to parallel then the speeds should be the
same and rays arrive in phase if the difference in distance is an
exact multiple of the reflected wavelength. That gives you the
grating equation:

lambda_r = D * sin(phi) / N

get it now?

I always did, I derived the above directly from your first attempt.
The bottom line is that a manufacturer ould make a grating with
an attached protractor and mark the scale with Angstroms
instead of degrees and you could read off the reflected
wavelength. He cannot mark it in nanoseconds for you to read
a time and the reading in Angstroms is _not_ the incident
wavelength since BaTh allows for a change of speed on
reflection.

You are the one who still isn't getting it even though you wrote
the equation yourself.

Exactly, the only evidence you have from any actual
obervations is for VDopppler alone. That's what I have
been pointing out all along. All the luminosity variations
are known to have other mundane explanations and
there is _no_ evidence for the existence of ADoppler
whatsoever.

No George, you aren't even trying to pass the test. ...

Correct, that is basic logic. If you want to prove ADoppler exists
you have to show that a result could _not_ be explained by an
alternative. The luminosity curves you have suggested can be
explained by intrinsic variability in Cepheids and by eclipses in
contact binaries so you have no proof.

...and it is pure coincidence that the shape of just about all variable star
curve just happens to match the BaTh prediction for simple orbiting stars?

It doesn't, you admitted already there are harmonics you
can't explain. However, that's an aside.

The point is that it _could_ be coincidence and your task is
to provide _proof_ that coincidence cannot explain it. Consider
what would happen if we found an eclipsing binary where the
velocity curves were at their maxima during the eclipse and
were perfect sine waves. That would show the orbits were
circular and the phase would indicate ADoppler. That's what
you need.

Bottom line is that you have offerred no scientific proof at all
for the existence of ADoppler.

Some of us physicists can put two and two together George....

Putting coincidences together is what laymen do Henry, you
could never be a physicist.

....
The movement BETWEEN photons continues for some
time.

Then each photon is moving at a mean different speed
from the speed of its parts which is nonsense, and if
you do a Fourier analysis you will find the modulation
of any wave will move at (c+v)/K when BaTh starts from
the assumption that it is (c+v). The result is self-
contradictory and therefore self-falsifying.

It isn't nonsense, George.

It is nonsense Henry, do a Fourier analysis if you
doubt me, you claimed you knew how to do that.

It is merely the mechanism of 'bunching', which you
illustrated yourself.

The bunching is valid and produces ADopppler as well as
VDoppler, but you will find it must apply at the same
level to pulses and cycles of a sine wave if you use a
Fourier analysis. That means K=1.

I keep telling you George, the intrinsic properties of individual photons must
be treated differently from those of the main 'bunching wave'.

And I keep telling you that if you do a Fourier analysis as
you claim you can, you will find that is impossible.

....
Here you go again...applying some kind of classical wave theory to light
particles.

BaTh as you have described it is a classical wave theory.

The group movenent of photons IS ballistic.

Yes, and ballistics is classical.

...not 'classical wave'....

It is "classical" because it treats the universe as deterministic
and being able to be exactly calculated using some combination
of waves, particles and so on. Post-classical theory treats all
events as being definable only in terms of probabilities and
the classical laws emerge only through the statistics of large
numbers of individual events.

Whether it uses waves or ballistics is neither here nor there
but your current formulation simply uses classical waves with
a modified speed equation. Perhaps it would be clearer if you
wrote down what you think is the BaTh equation for a propagating
wave.

What happens inside individual
photons is also ballistic but to a much smaller and limited extent.

Your "photons" are classical wavetrains, not point particles.

My best model is the 'serated bullet' one, where the serations represent a
'standing wave' or a helical path carved in space by something that rotates as
it moves.

It doesn't model the BaTh equations, the best match I could
get was the coil of wire moving along its axis. The back can
catch and pass through the front and the speed of each part
is independent of all other parts. That accurately reflects the
BaTh maths though it is harder to visualise.

Nope

lambda_r = D * sin(phi)

Incident light speed is prsent in 'lambda_r', George.

lambda_r has units of length Henry, don't be an idiot.

Lambda_r=lambda_i.(c+u/(c+v)), assuming light leaves the grating at c+u.

lambda_r is what is measured by the grating. That may imply
other things _IF_ you make _ASSUMPTIONS_ but lambda_r is
the only quantity that is actually measured.

see above.
I didn't think I would have to teach YOU geometry George.

We always agreed the geometry Henry. I didn't think I would
have to teach you algebra Henry but it seems I do.

Otherwise it is the same as
the classical one.

It measures reflected wavelength specifically but otherwise
is the same as the classical equation.

No it doesn't.

Yes it does Henry, your previous guesses were wrong.

It measures the time ...

D * sin(phi) / N does not have units of time Henry.

OK.... strictly speaking, it combines TIME and ABSOLUTE incoming wavelength
with the observed diffraction angle to calculate relative source speed .

No Henry, strictly speaking the angle depends _only_ on the reflected
wavelength. It does _not_ measure the incident wavelength. You might
try to infer that but you can only do so by making an _assumption_
about the change of speed on reflection.

That's OK. There is still a carrier frequency and a signal frequency.

Actually no, there is just a carrier and a 'local oscillator'
but the key point is that the same mixing technique works
as well for light as it does for audio and RF.

Some people have recently claimed that this is true.

This has been the basis of instruments for many years, you
are way out of date again.

My understanding is that only quite recently has light been mixed with very
short microwaves to create observable beats.

Laser combs have been in use for many years. I couldn't say
how long but I heard of their use a long time ago.

It still doesn't tell us much about photon 'frequency' because beating is
really a 'wavelength based' phenomenon.

You miss the point, beating is a linear effect based on addition of
waves, the envelope exhibits the difference frequecy but if you
analyse the composite with Fourier you only have the original
frequencies. Heterodyning produces actual signal power at the
sum and difference frequencies which requires _multiplication_
of the source waves. It is exactly the same technique used in
radio receivers and means that light behaves exactly the same
as radio. Anything I can do with a CW 1MHz transmitter can in
theory also be done with a mono-mode laser beam given fast
enough components

Same as for RF of course, a stream of phase-related photons.

Why not a periodic variation in photon density?

Variations in flux also apply to both.

How does one 'phase relate' photons anyway?

By making all the electrons in an antenna move in the
same direction at the same time, or by getting one
photon to prompt the emission of another in phase in
a laser.

Doesn't each electron emit a stream of photons as it accelerates George?

Since all the electrons move together under the influence
of the signal applied to the antenna, they emit in phase.

I don't agree. I say they each emit randomly but the RATE at which they emit is
governed by the signal.

No, the rate of arrival is just the intensity for light or the
carrier power for RF.

The carrier 'frequency' relates to varying photon
density.

No, it is intrinsic, a single photon still gets deflected by a
grating
by the same amount as the stream from which it was taken.

Come on!..., you don't know what happens to photons in a radio wave.

Exactly the same as light Henry.

Well, I suppose their are plenty of individual 'RF wavelength' photons produced
by thermal radiation and weak molecular bond transitions. ....but they don't
make a radio wave.

They do, thermal radiation can be detected as microwaves
but it has a broadband spectrum because the photons are not
phase related.

Photons in the visible light region and higher can be
detected individually, as you pointed out. Lots of similar photons all going in
the same direction make a maser or laser beam. Question: What happens to
individual photons inside a maser cavity?

They bounce off the ends and have a small probability of
escaping. As they pass some atoms, the promt the emission
of additional photons in phase.

....
Because individual photons are particle-like and what happens inside them
doesn't influence the bunching process at all.

Your model is classical waves, not point particles, but you
don't even need that, just apply Fourier to the macroscopic
sum of the photons which is a clasical wave travelling at
c+v.

If I were to assume that intrinsic photon oscillations interact, a fourier
combination would produce either white light or complete destructive
interference...I'm not sure which.

It can produce either constructive or destructive interference.
White light is broadband with a uniform spread of intrinsic
frequency and uncorrelated phases.

George, think of a pure sinusoidal RF signal. I say the observed wave effect is
just a result of photon density variation...or 'bunching'.

Visible light on the other hand is generally not like that at all... but
consists of identical photons whose energies add together to form 'beam
intensity'.

Visible light and radio are known to be the same thing differing
only in the frequency range. The fact that we can multiple laser
combs with incident light or with microwave signals using the
old heterodyne technique proves that.

Since each photon carries the same energy, bunching gives
changes of intensity. If what you say were true, a single photon
wouldn't have any preferred deflection angle when hitting a grating
but in fact in a low rate stream from a monochromatic source all
the photons get deflected by the same amount even when the
arrival rate is random. See the video or stills I cited.

In my model, an EM beam of a particular wavelength can be produced in two quite
different ways. It can be the result of either lots of identical photons or it
can reflect the bunching pattern formed in groups of random photons all
traveling in the same direction.

Consider the electron radiation from an RF antenna again. Each electron
experiences varying acceleration as it moves up and down the antenna...but all
are linked in phase. The radiation from each electron must be entirely random
but the overall flux density of radiation is still controlled by the signal.

Facts:

1) the "flux density" is controlled by the voltage fed to the
antenna,
not the frequency.

2) Each photon has the same frequency as the frequency of the AC
fed to the antenna.

The individual photons don't have to be 'phase-linked' in any way.

Then call it that, "matter" is an appropriate term.

It isn't just 'matter'. What is matter anyway?

Then call it the aether, whatever, "frame of reference" has an
entirely different meaning.

Not my EM FoR.

"Frame of reference" is purely mathematical, you are talking about
a combination of matter and an aether.


Well I wont dwell on this ...

Nor will I if you stop getting it wrong, it is only jargon,
not physics.

You're unusually stubborn today George.

You arethe one continually causing confusion by insisting on
being wrong Henry, why do you stubbornly persist in saying
"frame of reference" when you know it means something
completely different to what you are trying to describe?

I have now provided a clearer definition.

The accepted terms for what you are describing are matter and/or
an aether. Frame of reference is the coordiante scheme used to
describe the motion of the matter and the flow of the aether.

Frame of reference is mathematical only, matter is what
you mean.

Not this one...it's physical...

Still stubbornly insisting on being wrong Henry, why
don't you grow up a bit.

I have now provided a clearer definition.

You don't need a dfinition, both concepts have existing
terms to describe them. Stick with those accepted terms
Henry, I'll continue to point out where you get them wrong
for the benefit of lurkers who haven't seen this.

Also the reflected angle will not be exactly the
incident one.

Wrong again, since incident and reflected speeds are the
same, the angles are also the same.

But the reflected and incident speeds are NOT the same..

See above, both are c.

There is a subtle problem with reflection angles that we haven't considered at
all. I will get onto this soon.

No, no problem at all. The ratio of the angles depends on the
ratio of the speeds and since the speeds are the same, the
angles are the same.

Have another think about it George.

No need, we discussed this to death three years ago and you
discovered I was right, the diagram is still there.

We ignored a vital piece of information.

George, if you shoot from a moving car at an object lying at 45 degrees, what
is the direction of the bullet's CENTRAL AXIS when it hits?

I don't care, the thing measured in Sagnac is the time of flight
of the bullet.

...
If the object is also moving at your speed but perpendicularly away from the
road, how does that affect the angle at which the bullet will bounce of the
object (assuming specularly).

Do you mean vertically up from the Sagnac turntable? It would simply
go over the top of the detector. I'm not sure I understand your
question.

Photons are not 'little round balls'.

I think you will also find that the equation governing fringe shift turns out
to be similar to the aether theory one.

Nope, ballistic theory says there should be no fringe
shift whatsoever as we proved with your diagram and
my algebra:

http://www.briar.demon.co.uk/Henri/sagnac.gif

Remember that? You drew it and I just fixed a minor
error. The original might still be on your site somewhere
and the algebra is on Google.

it's wrong.

You drew it, it is correct and you agreed the algebra. Repeat
the analysis if you wish, it's only maths so you will get the
same result.

It's a lot more complicated.

You drew it and it correctly shows the full story as described
by ballistic theory. You also checked and agreed the equations
I worte relating to it which were in all honesty quite trivial. They
say there should be no fringe shift which is exactly what we
expect from ballistic theory, it predicts no shift.

Essentially what happens is that one beam moves around the ring at c+v/root2
and the other at c-v/root2 (wrt the non-rotating frame)...
The small difference in path length doesn't compensate for the difference in
travel times..

Do the algebra Henry, we showed the compensation
was exact including the "root2" factor.

Yes, I'm aware of what we did before.

Good, I can't be bothered digging it up again.

George