Why are the 'Fixed Stars' so FIXED?

"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light, you
WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

Normally, however fm would be lost because it is far from the frequencies
of interest.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines

correct.

but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

No.

A detector follows the 'envelope' of the modulated signal and
'demodulates' it, producing fm.

Following the envelope is essentially peak detection
with low pass filtering. Consider the effect of that
on the diagram for 50% modulation:

http://en.wikipedia.org/wiki/Amplitu...dulation_index

Would you not describe that as a time varying amplitude
or light intensity in the context? The amplitude of the
carrier varies at rate fm.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

George

"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light, you
WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

Normally, however fm would be lost because it is far from the frequencies
of interest.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines

correct.

but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

No.

A detector follows the 'envelope' of the modulated signal and
'demodulates' it, producing fm.

Following the envelope is essentially peak detection
with low pass filtering. Consider the effect of that
on the diagram for 50% modulation:

http://en.wikipedia.org/wiki/Amplitu...dulation_index

Would you not describe that as a time varying amplitude
or light intensity in the context? The amplitude of the
carrier varies at rate fm.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

George

"Henri Wilson" HW@.... wrote in message
...
On Sun, 6 May 2007 13:06:31 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
On Sat, 5 May 2007 08:50:53 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message

George, why don't you accept the fact that even today, nobody has the
faintest
idea of what a photon actually is.

Henry, why don't you just accept that photons from
a laser deflect by an angle determined by the colour
of the light and not the time between photon arrivals,
you did in a second post and disagreed in a third.

I distinguish between waves that are intrinsic to individual photons and
waves made from density distributions in large groups of photons.

So do I but the latter are merely statistical variations.

George, when signals are sent through optical fibres, how are they
modulated?

For telecomms, I believe usually AM. In fibre gyros, phase modulation.

You should know that the 'carrier light' can have a wide range of
wavelengths and still do the job.

Sure, it shows up as a bit of noise. What does that have to do
with you saying three different things on the same point in three
posts? how can we discuss this if you can't even keep your
story straight.

But we don't agree that the rate within a photon is far greater than
the rate BETWEEN photons.

The rate is fixed by your speed equalisation factor.

The inside of a photon has completely different properties from the
space between photons. Why should the two be the same?

Space has only one set of properties. Ballistic theory
says the speed is c+v tending towards c and that theory
applies to all the waves in your photon packet.

George, when you talk about the speed of anything you must always
provide
a reference. You should know that by now.

Are you denying ballistic theory says the speed is
c+v relative to the source?


You didn't mean that, I hope. You meant 'c relative to the source, c+v
relative
to the observer'.

Indeed, just a slip of the keyboard.

Are you denying it says
the speed is asymptotic to c/n relative to a medium
where n is the refractive index of that medium? I'm
just applying your theory consistently.

I'm not denying that....

Good, then what I said stands.

but strees that light entering such a medium might
never get even close to c/n (wrt the medium frame) before it passes right
through..

That's why I suggested you consider how a quarter-wave
plate works.

Ballistic theory says the speed of EM is INITIALLY c wrt its source and
c+v wrt
an object moving at -v wrt the source...

Refuted by De Sitter's argument.

Not refuted by DeSitters wrong argument.

The argument is correct.

... what happens to the light during
travel is not really part of the basic theory although we now suspect
that
it
experiences speed changes and speed unification....

If it isn't part of your theory, it fails, we should see
multiple images.

That idea was thrown out years ago.

No, it is still valid. If the theory doesn't include
some reduction of the speed difference between light
initially emitted at c+v and c-v then multiple images
must appear. That argument is and always will be valid.

Unification takes care of multiple imagery.

You need to learn to read more carefully, Henry, you
just said unification "is not really part of the basic
theory" so it doesn't take care of anything. Either it
is part of your theoryor it isn't, and de Sitter's
argument applies to the case where unification is _not_
part of the theory.

No star light seems to ever
overtakes other light....but there might be instances where it does.

There are many instances where it should, but it never
gets to within 0.1% of that, it is _never_ observed.

So are many orbit periods.

No orbital periods are more stable and don't show the
discontinuous phase changes of Cepheids.

There are plenty of complex orbit systems that would cause that
effect.

Nope, you can't gete a nice consistent value for years
with step discontinuities.

George, our own sun moves in a complex orbit around its barycentre with
all the
planets. Those small anomalies would show up in its brightness curve
50000
LYs
away.

Yes, and they would be smooth changes indicative
of Keplerian orbits. Cepheids show non-Keplerian
changes.

they don't.

Yes they do, study the subject before spouting.

....
The idea that individual detections "could barely
be seen above the noise" is ********, the detectors
are far less noisy than you imagine. That is obvious
in the stills.

They aren't photons. They're electrons..

Yes, and that is how PM tubes work (at least early
ones). The stills _are_ a converted PM detector and
if there was a high noise level it would be visible
in the photographs.

The theory says a photon (or several) knocks a single electron out of an
atom.

No, experiment says _one_ photon knocks _one_ electron
out of the surface. It takes some amount of energy to
free an electron, say W. If h.nu is less than W than
no electron gets released no matter how bright the
source so we know that "several" never happens. And if
h.nu W then one electron is liberated with a residual
kinetic energy of h.nu-W. If h.nu 2W a wave description
suggests more than one elctron could be liberated by a
single photon but again that doesn't happen.

The electron is then accelerated, causing an avalanche that is visually
recordable.

Right, and that's the part where I have shown you that the
noise levels are adequately low to be negligible in our
context.

The fact that the principle can be used to detect single photons is an
added bonus.


http://ophelia.princeton.edu/~page/single_photon.html

There is no PM in this experiment.

"The Hamamatsu camera is a remarkable device. In
essence, it has two successive micro-channel
plates followed by a CCD chip."

What do you think that is then?

It accelerates single electrons, emitting photon bursts. These are what
the
thing sees.

Yes, and in a photo-multiplier the first electron
is emitted by the photo-electric effect. The whole
amplification and detection process is identical.
It is in fact an actual PM camera with just the
front end removed so you can see the noise level
for yourself.

In any case, you aren't 'seeing' a single photon. You are merely verifying
that
an electron can be released by one.

'seeing' is amental process with our eyes acting as
input sensors, the PM tube is merely an extension of
that so in that sense we are 'seeing' single photons.

Your are wandering off the point though, each photon
gets deflected by a grating by an angle determined
solely by its intrinsic properties, not when the
next photon will arrive.

George, you keep telling me I have to match observed data.

A theory is required to be self-consistent as well as
matching the data.

If I assume K is 1, nothing matches.

The velocities do. The luminosity is then seen to be
intrinsic in eclipsing binaries and Cepheids. A small
value of 'extinction' distance is required for EF Dra
and the pulsars which is entirely consistent. Your
theory survives all these tests but in every case where
we can tell (there's no phase reference for Cepheids)
only VDoppler can be seen.

George, if it weren't for the fact that a great many brightness curves can
be
matched with BaTh, ...

Sorry Henry, you can't match any without making your model
self-contradictory. You _can_ match the velocity curves
but not luminosity.

.. I would take the easy way out and probably agree with you.
However, since logic tells us that there is no mechanism outside of
fairyland
which would cause all starlight in the universe to travel towards little
planet
Earth at precisely c, and since I CAN match brightness curves very nicely,

No you can't, all you can match is curves of less than 0.002
magnitude variation, max.

I
will prefer to continue along my present very interesting and fruitful
path.

Fair enough, I'll continue to dismiss it and point out
the truth to anyone following the thread until you make
it consistent.

If I assume it has a value of maybe 10000,
then everything falls into place, I can match hundreds of brightness
curves in
phase and magnitude with velocity curves.

But it is then self-contradictory so fails to be a theory
in the first place.

It isn't. It can have a value of 10000 ..

Nope, that requires the light to travel at both c+v and
(c+v)/10000 at the same time, it is self-contradictory.

George, this is how exepriment physics operates. If K is not = 1, then
all
data
is matched. What is the logical conclusion?

Without K=1 you cannot match simple Doppler measurements
in the lab and K1 conflicts with c+v for the speed, it
is self-contradictory so proves itself wrong.

I now consider that Labs create and constitute their own strong EM FoRs.

An "FoR" is a mathematical coordinate system with no
physical existence.

that uses frequency can equally well be written
using speed and wavelength. You really need to find
out what your equation is before you make a bigger
fool of yourself.

George, I can say whatever I like and you can't prove me wrong.

Yes I can if what you say conflicts with what you say,
one or the other is wrong. Either you know frequency is
the independent variable in the equation or you don't
know what the equation is, both cannot be true.

Nobody has
moved a grating in remote space ...

Itrrelevant, what equation for aa grating deflection
angle is derived from the BaTh basic equations by pure
maths?

I will soon produce the relevant diagram for htis.

Don't waste your time, just show your mathematical
derivation of the equation from c+v.

It should be pretty obvious.

It should, in fact it's a problem that you should be
able to do in a few minutes, but your incapable of
even the simplest algebra from what I have seen.

THE BLOODY BRIGHTNESS PEAK IS EXACTLY IN PHASE WITH THE CENTRE OF THE
ECLIPSE.

Yes, but the observed velocity peak is exactly between
the eclipses, and the period of the orbit is double
the period of the eclipses giving a 45 degree error.

Oh, Ok. I wasn't looking at that.

OK, you need to have a more detailed look. It isn't
trivial.

No, it certainly isn't.
I just hadn't gotten around to it.

Right, you just faked the result and got caught out.

Yes that's interesting...and backs up my theory that unification is
pretty
quick near short period binaries and also that K 1.
It means there is still enough ADoppler to account for the brightness
variation
although the individual photons are essentially VDoppler shifted.

I doubt it, but remember the eclipses will fully
explain the luminosity anyway so you don't need
to worry about matching that curve at all, only
the velocity curves. The spectral shift is the
same no matter if part of the star is hidden as
long as there is enough light to measure.

The curves don't really tell us much because there are only a few points
to go
on.

They tell us where the peaks are and that phase is what we
need to know.

Which is the BaTh prediction.

Wrong. If you had used you program instead of faking
your results, you would have found that yourself.

Well you can see a better curve now.

http://www.users.bigpond.com/hewn/efdra.jpg

As I write it still matches the luminosity instead of
the velocities.

Yes.

Pointless then the luminosity is dominated by the
two eclipses. Do one matching the velocity curves.

K is obviously large for close binaries...but not so large for cepheids.

K is 1, period.

Right, the 'wavelength' of the photons is what
determines the grating deflection angle.

...and that 'wavelength' cannot possibly change just because the GRATING
moves.

I have explained several times why BaTh says it
_can_ change. You need to do the derivation to
find out if it predicts that it does.

BaTh says the difraction angles are sensitive to 'wavecrest arrival rate'.

No it doesn't, it says the speed is c+v initially and
that approaches c/n according to the formula

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

To get from there to an equation will take you some work.

I will illustrate the principle today if I get a chance.

Just show me the equation and stop guessing.

The FREQUENCY of wavecrest arrival is what the BaTh uses.

You can't seriously be trying to tell me you would
put 1Hz into the BaTh equation for the grating
deflection, are you? I certainly gave you credit
for more understanding than that. The grating
angle depends on the colour of the light, not how
many photons per second arrive.

That's OK for light....but not for generated radio waves.

Both are EM, any theory must be equally aplplicable to
both.

But George, you are not distinguishing between a beam of light made from a
large number of identical photons, all moving at the same speed, and a
generated radio signal made up of intelligently bunched groupings of any
old
photons.

There is nothing to distinguish, a mono-mode laser signal
is a generated signal exactly the same as the RF signal
but at a higher frequency. Early radio receivers used a
"heterodyne" technique to improve tuning, high resolution
spectroscopy does exactly the same by heterodyning the
starlight with a laser and measuring the beat frequency
with an RF receiver.

I'm saying the radio waves use 'photon density' variations, whereas light
rays
use intrinsic photon properties.

You can't realy believe that a constant RF signal lasting ten years is
made of
one single photon.

No, nor do I believe a mono-mode laser running for ten
years emits a single photon.

Well what's you model for this?

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

So what's the difference George? Are you going to offer any suggestions?

None, both consist of a flux of many photons.

What's wrong with my above model?

It tries to explain a difference that doesn't exist.

Tell me, what is the relationship between an constant RF sine wave and a
photon?

Same as for a mono-mode laser, bz has told you already
so I won't repeat it.

BZ knows nothing....but he tries....

He knows vastly more than you, but like everyone else
his answers are over your head because you haven't
spent the time learning the basics. Tools like Fourier
analysis are essential if you are going to follow more
complex theories.

Yep, it also mean ADoppler is non-existent for binaries,
the light changes to speed c within 4.6 microns of leaving
the star's surface ;-)

That's c wrt the star George.

It is c wrt to the material with which it is interacting
to cause the speed change Henry, otherwise you cannot
transfer the energy and momentum to maintain conservation.

You can't assume it is 'material'. Just call it a 'local EM FoR'.

Why would I want to look stupid, you don't transfer
momentum to a coordinate system.

For contact binaries, it appears that such a frame is defined by the
barycentre
of the pair.

Garbage, the frame is chosen by whoever does the calculations.

However, I agree, it also appears to quite rapidly approach 'c' wrt the
BARYCENTRE of the pair in the case of pulsars and short period binaries.

This again raises the question, "how and why does unification rate
depend
on
period?"

I have answered that before in some detail twice but
it is a subtle point and you didn't really follow it.
Basically it shows the theory is unlikely to be true
because it requires a remarkable coincidence between
your pitch factor and the peak orbital acceleration.

I don't have a definite view on this yet.

I know, you wont be able to follow the argument. You
might start to see it if you could draw a cross-section
of a binary system and plot 'isobars' of extinction
distance but I doubt even that would switch the light
bulb on.

De Sitter was wrong.. face it George.

He was right, or you wouldn't need extinction.

I can live with extinction. De Sitter couldn't.

He didn't have to, it had to be invented as a result
of his falsification of Ritz's theory.

...and no other experiment refutes the BaTh.

Sagnac and Shapiro do.

Other factors are involved.

As with De Sitter, they falsify BaTh as it stands. If
you want to come up with a new alternative then maybe
will have other problems, but as it stands at the
moment Sagnac and Shapiro both independently falsify
BaTh.

I have already suggested that BaTh applies 100% only in genuinely empty
space.

For Ritz's theory that would be true, speed equalisation
like a refractive index requires material.

I am also of the opinion that local EM FoRs are present wherever matter or
fields exist.

Still showing your ignorance Henry, a frame of reference
is purely a mathematical device for assigning coordinates.

It is quite possible that there may be a compromise theory that might
explain
the intricacies of starlight movement and still accommodate some aspects
of
Einstein's modified aether theory.

I sense that you may be thinking along similar lines.

No, I'm thinking you have been corrected on most of the
string of stupid errors you made many times before and I
wonder how you can persist in making a fool of yourself
over and over again without leaving the group to avoid
further embarrassment. It's just one of life's little
mysteries.

George

"George Dishman" wrote in
:


"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light,
you WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

You may be correct. That math would seem to support it.

I would have sworn that the modulating signal also WOULD appear if it were
not filtered out. And it normally is because it is normally audio and the
carrier is RF.

For instance, in the case of mixing two signals that are close to the same
frequency, as in a hetrodyne receiver, you get the sum, the difference and
both f1 and f2.

Normally, however fm would be lost because it is far from the
frequencies of interest.

My understanding is that the stream contains a mixture
of three frequencies of photons and if you have the
resolving power in the grating, you get three lines

correct.

but a lower resolution will cause the lines to overlap
and the interference then causes the time varying
intensity.

No.

A detector follows the 'envelope' of the modulated signal and
'demodulates' it, producing fm.

Following the envelope is essentially peak detection
with low pass filtering. Consider the effect of that
on the diagram for 50% modulation:

http://en.wikipedia.org/wiki/Amplitu...dulation_index

Would you not describe that as a time varying amplitude
or light intensity in the context? The amplitude of the
carrier varies at rate fm.

2 points for you.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

Looks like either the designations have changed since I took my exams
(first class radio telephone, 2nd class radio telegraph, amateur extra
class) or my memory has played a nasty trick on me.

A check of my handbooks shows that memory is the offending agent.
A3 is double side band, A3A single, reduced carrier. A3J single,
suppressed carrier. I notice that A3E is the current designation for
double sideband with full carrier.

Well, hopefully, it won't be the last time I am wrong about something.




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

"bz" wrote in message
98.139...
"George Dishman" wrote in
:
"bz" wrote in message
98.139...
"George Dishman" wrote in news:f1mmtj$tr3$1
@news.freedom2surf.net:

Suppose the unmodulated light has a frequency of fc.
If you fire the modulated light at a grating There
are two obvious possibilities, either you get a line
with time varying intensity at an angle corresponding
to fc, or you get a signal which has a carrier fc and
two sidebands at +/- fm

fc
|
fc-fm | fc+fm
______|____|____|______

and each frequency produces a line of constant intensity.
Either way, you don't get any power at the angle
corresponding to fm itself.

I am assuming fully linear mixing with modulation
index 1, no sideband or carrier suppression.

IF the amplifiers following the mixer were flat from DC through light,
you WOULD also have output at fm.

sin(a)*sin(b) = (cos(a-b) - cos(a+b))/2

So

sin(fc.t)*(1+M*sin(fm.t)) =
sin(fc.t) + M/2*(cos((fc-fm).t) - cos(((fc+fm).t))

There is no component at fm, only the three I listed.

You may be correct. That math would seem to support it.

I would have sworn that the modulating signal also WOULD appear if it were
not filtered out. And it normally is because it is normally audio and the
carrier is RF.

The carrier apears because of the "(1+" term and likewise
the modulating signal will slip through if there is a DC
bias on the RF. In practical terms there usually will be
in a physical circuit but not in the case of light incident
on an optical modulator. As you say, it always gets filtered
out.

For instance, in the case of mixing two signals that are close to the same
frequency, as in a hetrodyne receiver, you get the sum, the difference and
both f1 and f2.

For a perfect four-quadrant multiplier with no DC offsets,
you only get sum and difference. f1 gets through if there
is DC on the f2 signal and vice versa.

[all the above assumes A3A modulation commonly called AM or amplitude
modulation].

A3A is single sideband, suppressed carrier. I was
describing A3 mode, both sidebands, M1 and no carrier
suppression.

Looks like either the designations have changed since I took my exams
(first class radio telephone, 2nd class radio telegraph, amateur extra
class) or my memory has played a nasty trick on me.

A check of my handbooks shows that memory is the offending agent.
A3 is double side band, A3A single, reduced carrier. A3J single,
suppressed carrier. I notice that A3E is the current designation for
double sideband with full carrier.

I was never a ham though I had some friends who were,
my interest was always in the digital side. I Googled
and randomly got this page:

http://jproc.ca/rrp/coverdale_ddr5k.html

Well, hopefully, it won't be the last time I am wrong about something.

No problem, I've learned more than you from this :-)

George

"George Dishman" wrote in news:f1o0ir$30m$1
@news.freedom2surf.net:

I was never a ham though I had some friends who were,
my interest was always in the digital side. I Googled
and randomly got this page:

http://jproc.ca/rrp/coverdale_ddr5k.html

For something a bit more 'state of the art'
http://www.elecraft.com/K3/K3_Data%20Sheet_rev06.pdf


Well, hopefully, it won't be the last time I am wrong about something.

No problem, I've learned more than you from this :-)

Wanna bet?




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

bz wrote:

For something a bit more 'state of the art'
http://www.elecraft.com/K3/K3_Data%20Sheet_rev06.pdf

What kind of funky microphone connector is on the front panel
of this baby? Looks non-standard.

Steve

On 7 May 2007 09:55:54 -0700, George Dishman wrote:


"Henri Wilson" HW@.... wrote in message
.. .


Unification takes care of multiple imagery.

You need to learn to read more carefully, Henry, you
just said unification "is not really part of the basic
theory" so it doesn't take care of anything. Either it
is part of your theoryor it isn't, and de Sitter's
argument applies to the case where unification is _not_
part of the theory.

No star light seems to ever
overtakes other light....but there might be instances where it does.

There are many instances where it should, but it never
gets to within 0.1% of that, it is _never_ observed.

I don't know where you got that figure from.



The theory says a photon (or several) knocks a single electron out of an
atom.

No, experiment says _one_ photon knocks _one_ electron
out of the surface. It takes some amount of energy to
free an electron, say W. If h.nu is less than W than
no electron gets released no matter how bright the
source so we know that "several" never happens. And if
h.nu W then one electron is liberated with a residual
kinetic energy of h.nu-W. If h.nu 2W a wave description
suggests more than one elctron could be liberated by a
single photon but again that doesn't happen.

The electron is then accelerated, causing an avalanche that is visually
recordable.

Right, and that's the part where I have shown you that the
noise levels are adequately low to be negligible in our
context.




It accelerates single electrons, emitting photon bursts. These are what
the
thing sees.

Yes, and in a photo-multiplier the first electron
is emitted by the photo-electric effect. The whole
amplification and detection process is identical.
It is in fact an actual PM camera with just the
front end removed so you can see the noise level
for yourself.



The velocities do. The luminosity is then seen to be
intrinsic in eclipsing binaries and Cepheids. A small
value of 'extinction' distance is required for EF Dra
and the pulsars which is entirely consistent. Your
theory survives all these tests but in every case where
we can tell (there's no phase reference for Cepheids)
only VDoppler can be seen.

George, if it weren't for the fact that a great many brightness curves can
be
matched with BaTh, ...

Sorry Henry, you can't match any without making your model
self-contradictory. You _can_ match the velocity curves
but not luminosity.

I can easily match both George.

.. I would take the easy way out and probably agree with you.
However, since logic tells us that there is no mechanism outside of
fairyland
which would cause all starlight in the universe to travel towards little
planet
Earth at precisely c, and since I CAN match brightness curves very nicely,

No you can't, all you can match is curves of less than 0.002
magnitude variation, max.

George this is a plainly ridiculous claim. If you could set up your own program
(too hard, no doubt) you would soon see that (log) magnitude variations of
three or more can easily be achieved before peaks appear in the brightness
curves.

will prefer to continue along my present very interesting and fruitful
path.

Fair enough, I'll continue to dismiss it and point out
the truth to anyone following the thread until you make
it consistent.

Well I have now solved Sagnac.,,so that will please you even more...

If I assume it has a value of maybe 10000,
then everything falls into place, I can match hundreds of brightness
curves in
phase and magnitude with velocity curves.

But it is then self-contradictory so fails to be a theory
in the first place.

It isn't. It can have a value of 10000 ..

Nope, that requires the light to travel at both c+v and
(c+v)/10000 at the same time, it is self-contradictory.

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.

George, this is how exepriment physics operates. If K is not = 1, then
all
data
is matched. What is the logical conclusion?

Without K=1 you cannot match simple Doppler measurements
in the lab and K1 conflicts with c+v for the speed, it
is self-contradictory so proves itself wrong.

I now consider that Labs create and constitute their own strong EM FoRs.

An "FoR" is a mathematical coordinate system with no
physical existence.

An EM FoR is one of limited size that sets light speed somewhat loosely, within
itself.


Yes I can if what you say conflicts with what you say,
one or the other is wrong. Either you know frequency is
the independent variable in the equation or you don't
know what the equation is, both cannot be true.

Nobody has
moved a grating in remote space ...

Itrrelevant, what equation for aa grating deflection
angle is derived from the BaTh basic equations by pure
maths?

I will soon produce the relevant diagram for htis.

Don't waste your time, just show your mathematical
derivation of the equation from c+v.


It should be pretty obvious.

It should, in fact it's a problem that you should be
able to do in a few minutes, but your incapable of
even the simplest algebra from what I have seen.

Well you've seen it now.
http://www.users.bigpond.com/hewn/bathgrating.jpg

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


Oh, Ok. I wasn't looking at that.

OK, you need to have a more detailed look. It isn't
trivial.

No, it certainly isn't.
I just hadn't gotten around to it.

Right, you just faked the result and got caught out.

I did not fake anything George. I just draw a rough curve to show you the basic
shape of the brightness curve of one member. I can't match it exactly because
most of it is hidden.



The curves don't really tell us much because there are only a few points
to go
on.

They tell us where the peaks are and that phase is what we
need to know.

...and it all fits nicely....

Which is the BaTh prediction.

Wrong. If you had used you program instead of faking
your results, you would have found that yourself.

Well you can see a better curve now.

http://www.users.bigpond.com/hewn/efdra.jpg

As I write it still matches the luminosity instead of
the velocities.

Yes.

Pointless then the luminosity is dominated by the
two eclipses. Do one matching the velocity curves.

The velocity curves are basically VDoppler..because the individual photons very
rapidly become stabilized. The movement BETWEEN photons continues for some
time.

K is obviously large for close binaries...but not so large for cepheids.

K is 1, period.

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

Right, the 'wavelength' of the photons is what
determines the grating deflection angle.

...and that 'wavelength' cannot possibly change just because the GRATING
moves.

I have explained several times why BaTh says it
_can_ change. You need to do the derivation to
find out if it predicts that it does.

BaTh says the difraction angles are sensitive to 'wavecrest arrival rate'.

No it doesn't, it says the speed is c+v initially and
that approaches c/n according to the formula

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

To get from there to an equation will take you some work.

I will illustrate the principle today if I get a chance.

Just show me the equation and stop guessing.

http://www.users.bigpond.com/hewn/bathgrating.jpg

Note, light speed is included in the BaTh equation. Otherwise it is the same as
the classical one.



But George, you are not distinguishing between a beam of light made from a
large number of identical photons, all moving at the same speed, and a
generated radio signal made up of intelligently bunched groupings of any
old
photons.

There is nothing to distinguish, a mono-mode laser signal
is a generated signal exactly the same as the RF signal
but at a higher frequency. Early radio receivers used a
"heterodyne" technique to improve tuning, high resolution
spectroscopy does exactly the same by heterodyning the
starlight with a laser and measuring the beat frequency
with an RF receiver.

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


You can't realy believe that a constant RF signal lasting ten years is
made of
one single photon.

No, nor do I believe a mono-mode laser running for ten
years emits a single photon.

Well what's you model for this?

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

Why not a periodic variation in photon density?

How does one 'phase relate' photons anyway?

So what's the difference George? Are you going to offer any suggestions?

None, both consist of a flux of many photons.

What's wrong with my above model?

It tries to explain a difference that doesn't exist.

Tell me, what is the relationship between an constant RF sine wave and a
photon?

Same as for a mono-mode laser, bz has told you already
so I won't repeat it.

BZ knows nothing....but he tries....

He knows vastly more than you, but like everyone else
his answers are over your head because you haven't
spent the time learning the basics. Tools like Fourier
analysis are essential if you are going to follow more
complex theories.

George, I spent years analysing sine waves that make different musical
instrument sounds. I know all about it.

Yep, it also mean ADoppler is non-existent for binaries,
the light changes to speed c within 4.6 microns of leaving
the star's surface ;-)

That's c wrt the star George.

It is c wrt to the material with which it is interacting
to cause the speed change Henry, otherwise you cannot
transfer the energy and momentum to maintain conservation.

You can't assume it is 'material'. Just call it a 'local EM FoR'.

Why would I want to look stupid, you don't transfer
momentum to a coordinate system.

A local EM FoR is more than a cooordinate system. It contains matter and fields
that define a macroscopic reference for velocity.

For contact binaries, it appears that such a frame is defined by the
barycentre
of the pair.

Garbage, the frame is chosen by whoever does the calculations.

Well I wont dwell on this ...




...and no other experiment refutes the BaTh.

Sagnac and Shapiro do.

Other factors are involved.

As with De Sitter, they falsify BaTh as it stands. If
you want to come up with a new alternative then maybe
will have other problems, but as it stands at the
moment Sagnac and Shapiro both independently falsify
BaTh.

I have already suggested that BaTh applies 100% only in genuinely empty
space.

For Ritz's theory that would be true, speed equalisation
like a refractive index requires material.

I am also of the opinion that local EM FoRs are present wherever matter or
fields exist.

Still showing your ignorance Henry, a frame of reference
is purely a mathematical device for assigning coordinates.

I didn't say 'FoR'. I said an 'EM FoR'.
It's a physical entity not a mathematical one.

It is quite possible that there may be a compromise theory that might
explain
the intricacies of starlight movement and still accommodate some aspects
of
Einstein's modified aether theory.

I sense that you may be thinking along similar lines.

No, I'm thinking you have been corrected on most of the
string of stupid errors you made many times before and I
wonder how you can persist in making a fool of yourself
over and over again without leaving the group to avoid
further embarrassment. It's just one of life's little
mysteries.

Well I have now solved the Sagnac mystery.

As you know, specular reflection can be regarded a diffraction process with
reinforcement occuring at exactly the angle of incidence.
Now, you will see from my grating diagram that if the mirror is moving wrt the
source, the incident speed is c+v BUT THE REFLECTED SPEED IS probably 'c' or
thereabouts, wrt the mirror. Also the reflected angle will not be exactly the
incident one.
Applying this to Sagnac, it is easy to see that one beam ends up moving a lot
more slowly that the other. Hence the fringe shift.

The BaTh wins again.

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



George




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

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

On Mon, 7 May 2007 11:17:54 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .

see: www.users.bigpond.com/hewn/bathgrating.jpg

Well done Henry. So your equation is

lambda_i * (c+u)
sin(phi) = ---------------
D * (c+v)

where lambda_i is the wavelength of the _incident_
light.

The wavelength of the reflected light, lambda_r, is
given by

lambda_r c+u
-------- = ---
lambda_i c+v

So your equation can also be written

lambda_r
sin(phi) = --------
D

You have been claiming that the speed didn't appear in
the equation and that wavelength couldn't change. One
or the other is wrong. You also claimed the formula
used frequency instead of wavelength but that too isn't
true. Naturally you can replace the wavelength by speed
over frequency but that just reintroduces speed in the
equation.

Desperate again George?

Lambda_i is absolute and all we need.

Lambda_r doesn't enter into this.

The equation uses points of equal phase to calculate the angle of the wavefront
of the diffracted beam.

Let's assume that u =0, ie., the reflected light moves at c wrt the GRATING.

The result is as I said: Sin(phi)=D/lambda.(c/(c+v)), for 1st order
diffraction.
Speed is included in the equation....so the BaTh explains what is observed.

SR does not.


The lesson Henry, is to work out the equation before you
start telling people what it contains.

The BaTh wins again.
The BaTh also explains sagnac.
The BaTh wins yet again.

George




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

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

(Steve Pope) wrote in news:f1o818$rcd$1
@blue.rahul.net:

bz wrote:

For something a bit more 'state of the art'
http://www.elecraft.com/K3/K3_Data%20Sheet_rev06.pdf

What kind of funky microphone connector is on the front panel
of this baby? Looks non-standard.

Elecraft uses that connector on the K2 and the K3. It allows use of a
headphone with boom mike, ptt and some control lines, if desired. Seems to
work well enough.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.