Why are the 'Fixed Stars' so FIXED?

Henri Wilson wrote:
On Sun, 02 Sep 2007 21:14:39 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
sorOn Sat, 01 Sep 2007 03:17:23 +0200, "Paul B. Andersen"
wrote:

The 'nonsensical' theory is Planck's black body radiation law.
Which is so well confirmed that not even you will question it.
Or do you? :-)
It is reasonably well confirmed.
It is extremely well confirmed.
So why did you call it nonsensical?

It is 100% confirmed when a 'black body' is defined as an emitter that obeys
Planck's curve.

Is that the source of your confusion?

Impressive what you don't know, Henri. :-)

http://www.yourdictionary.com/ahd/b/b0295200.htm
"A theoretically perfect absorber of all incident radiation."

A practical way to make a near perfect black 'body' is to
have a cavity with a hole in it. The hole will 'absorb'
all the radiation that hits it, and will thus be black
according to the definition above.

If you heat the inside of the cavity (oven), the hole will
radiate 'black body radiation'.
http://rocinante.colorado.edu/~pja/a.../lecture06.pdf
This is the traditional way of producing BB radiation in a lab.

The spectrum of this radiation is exactly as predicted by Planck's law.

What isn't confirmed is whether or not the
average cepheid has a black body spectrum. Nor has it been confirmed that its
spectrum would remain black body if it went 'huff puff' all day long.
As my calculation shows, the observed light curves in K and V from l Carinae
are consistent with a black body spectrum.

....that might be true if the observations wer made close up...but the datat
you have is only Willusory.

What we have is the measured light curve in K and V, the measured
radius variation and the measured temperature variation.
All the measured data are consistent, the light curves in K
and V are exactly as they should be according to conventional theory.

A coincidence? :-)

What you have is a theory which is utterly unable to predict
the measured data. It predicts 'wilusions' which are not observed.

But since you find the verbal description above unconvincing,
(you didn't understand it, did you?) let's do the calculation properly.

As the primary, measured data, I will use the temperature in fig. 4.3 in:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf
That's a bloody PhD thesis....Do you really think a Phd student is going to
stand up and announce that the whole of astronomy is bull**** because it is
based on all starlight traveling to little planet Earth at precisely speed 'c'?
Of course not Tussellad...
In other words, you won't expect a Phd student to be a crank. :-)

...no, just someone who does what his supervisor tells him to do...and his
supervisor is bound to be an inbred member of the incestuous physics
establishment that worships the Einstein god.

You have to utterly ignorant of physics and astronomy to be open minded
enough to claim that "the whole of astronomy is bull****", right? :-)

I'm sorry Paul...but ignoring Willusions is on par with assuming everything we
currently see in the cosmos is happening right this instant.
For some obscure reason, the obvious is often very hard to identify.....

Your desperate claim that we cannot measure the temperature of stars
by analysing the absorption lines is not very convincing.


...but obviously, the 'obvious' hasn't reached Norway yet.

All the information used there is willusory...the paper is full of speculative
remarks made by a poor bugger who obviously trying to match one lot of nonsense
with more nonsense...

....so you are already on the wrong track....
Hardly a convincing argumentation for why the temperature curve
in fig. 4.3 must be wrong. :-)

And note one very important issue:
The temperature is derived from the absorption lines, and NOT
from the black body spectrum, which would have made my calculations
somewhat circular.

...that's not what I read.
Most of the lines are emission....lots of Fe, etc.,...
Anyway it probably makes little difference.

Your ignorance shows again, Henri.
All the 'spectral lines' in tab 4.1 are absorption lines.

and the radius curve in fig 3 in:
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf
Interferometry relies on all starlight moving at constant c wrt Earth.
Since that is not true, the technique can best be regarded as highly suspect.
The published cuvre is lousy anyway...nothing like a best fit....
Don't be such a blatant idiot, Henri.
If the BaTh predicts that it is possible to use a telescope
to measure the angular distance between two stars, then it also predicts
that it is possible to use an interferometer to measure the angular
diameter of a star. This is simple geometry, Henri. The geometry
is the same in both cases, and it doesn't matter what the speed
of light is.

Not so Paul. If the star is spinning..as it no doubt is..light from each side
will move at diffferent speeds towards earth..causing a willusory phase
difference.

This is mindless babble, Henri.
The "phase difference of light from different sides of the star"
is a meaningless concept - there is no such thing!
To compare the 'phase' of the light (noise!)
from two uncorrelated noise sources is meaningless!
What the hell should you compare?
And why do you think it is done anywhere?

Anyway, how can interferometry work at all?

It is indeed apparent that you don't understand how!

It requires coherent light and a split beam....or rather a split photon.....
Even you wouldn't claim that the same photon is emitted from both sides of the
star.

Should I laugh or should I cry?
I think I will laugh - a resigned laugh - this idiot is beyond reason.

Read my other posting.

Either the BaTh predicts that light behaves according
to the optical laws we know apply, or it doesn't.
And you are not insisting that it doesn't, are you? :-)
If you do, the very fact that telescopes work falsifies the BaTh.

Interferometry apparently detects some kind of change....but it sure ain't the
star's radius....


The fit is so good that one could think I have cheated.
But I haven't. You can check the calculations yourself,
if you don't believe me.
The fit is good simply because Planck's black body radiation law
is correct, and a Cepheid is what it is known to be - a pulsating star.
It is highly possible that some stars DO pulsate.
I take this remark as an admission that you accept
that l Carinae is a pulsating star, and the reason for
why the K and V light curves are as they are simply is
Planck's law.

The two curves can be produced using BaTh. They differ only in eccentricity and
yaw angle..as well as phase.

So why did you call this explanation 'nonsensical'?

Because it is all based on willusroy data..

The fact that their
brigthness curves match those of stars in an elliptical orbit of e ~ 0.15-.25
and yaw angle -50-70 is purely coincidental.

Do you still find Planck's blackbody radiation law nonsensical, Henri?
Planck's law was empirically derived for what is assumed to be a perfect black
body.
Planck's law can be derived from the assumption (postulate) that
the energy of the oscillators in the radiating body is quantized.

Stars vary considerably and not many fit that curve well at all.
Nonsense.
All stellar spectra are black body spectra with absorption and
emission lines due to the fact that the black body radiation
from the stellar photosphere has to go through the higher layers
of the stellar atmosphere.

...Now even your colleagues will correct you on that one Paul..

Henri, zip your mouth, your ignorance shows.

The spectrum is given primarily by the surface temperature of
the star, and varies remarkably little between stars with equal
temperatures.

...but not variables....

- but not variables - what?

We can sum it up thus:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.

dream on....

...is what you say when you don't know what to say, right?

Whether you like it or not, these are indisputable FACTS:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.

If you still haven't fathomed that, you better read again
my original posting where I prove the above to be facts.

As the primary, measured data, I will use the temperature in fig. 4.3 in:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf
and the radius curve in fig 3 in:
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The question is:
What will the K (2.2u) and V (0.5u) light curves be
according to Planck's blackbody radiation law?

The result is shown in the table below. Here a
Int = the surface radiation intensity relative to the intensity at phase 0.
Lum = the luminosity (intensity*area) relative to the luminosity at phase 0.
Mag = the magnitude relative to the magnitude at phase 0.

The Intensity is calculated from Planck's black body radiation law.
Planck(T,lambda). (Look it up if you don't know it.)
K Int = Planck(T,2.2u)/Planck(5600,2.2u)
V int = Planck(T,0.5u)/Planck(5600,0.5u)

Phase: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Temp: 5600 5550 5250 5050 4950 4900 4850 4950 5050 5400
Radius: 2.78 2.88 3.10 3.20 3.20 3.15 3.08 2.95 2.77 2.62
Area: 1.00 1.07 1.24 1.32 1.32 1.28 1.23 1.13 0.99 0.89
K Int: 1.00 0.98 0.89 0.84 0.81 0.79 0.78 0.81 0.84 0.94
K lum: 1.00 1.06 1.11 1.11 1.07 1.02 0.95 0.91 0.83 0.83
K mag: 0.00 -0.06 -0.12 -0.11 -0.07 -0.02 0.05 0.11 0.20 0.20
V Int: 1.00 0.95 0.71 0.57 0.51 0.48 0.45 0.51 0.57 0.83
V lum: 1.00 1.02 0.88 0.76 0.67 0.61 0.55 0.57 0.57 0.73
V mag: 0.00 -0.03 0.14 0.31 0.43 0.53 0.64 0.61 0.62 0.34

Compare K mag and V mag to the curves in fig.1 in
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The fit is so good that one could think I have cheated.
But I haven't. You can check the calculations yourself,
if you don't believe me.
The fit is good simply because Planck's black body radiation law
is correct, and a Cepheid is what it is known to be - a pulsating star.



Paul

On Thu, 06 Sep 2007 00:10:30 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Sun, 02 Sep 2007 21:14:39 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
sorOn Sat, 01 Sep 2007 03:17:23 +0200, "Paul B. Andersen"
wrote:

The 'nonsensical' theory is Planck's black body radiation law.
Which is so well confirmed that not even you will question it.
Or do you? :-)
It is reasonably well confirmed.
It is extremely well confirmed.
So why did you call it nonsensical?

It is 100% confirmed when a 'black body' is defined as an emitter that obeys
Planck's curve.

Is that the source of your confusion?

Impressive what you don't know, Henri. :-)

http://www.yourdictionary.com/ahd/b/b0295200.htm
"A theoretically perfect absorber of all incident radiation."

A practical way to make a near perfect black 'body' is to
have a cavity with a hole in it. The hole will 'absorb'
all the radiation that hits it, and will thus be black
according to the definition above.

If you heat the inside of the cavity (oven), the hole will
radiate 'black body radiation'.
http://rocinante.colorado.edu/~pja/a.../lecture06.pdf
This is the traditional way of producing BB radiation in a lab.

The spectrum of this radiation is exactly as predicted by Planck's law.

I'm quite aware of all that.

What isn't confirmed is whether or not the
average cepheid has a black body spectrum. Nor has it been confirmed that its
spectrum would remain black body if it went 'huff puff' all day long.
As my calculation shows, the observed light curves in K and V from l Carinae
are consistent with a black body spectrum.

....that might be true if the observations wer made close up...but the datat
you have is only Willusory.

What we have is the measured light curve in K and V, the measured
radius variation and the measured temperature variation.
All the measured data are consistent, the light curves in K
and V are exactly as they should be according to conventional theory.

A coincidence? :-)

No coincidence.
The three wilusory measurements merely indicate the same variation in light
speed from the star.

What you have is a theory which is utterly unable to predict
the measured data. It predicts 'wilusions' which are not observed.

Is everyone in Norway this stupid?



You have to utterly ignorant of physics and astronomy to be open minded
enough to claim that "the whole of astronomy is bull****", right? :-)

I'm sorry Paul...but ignoring Willusions is on par with assuming everything we
currently see in the cosmos is happening right this instant.
For some obscure reason, the obvious is often very hard to identify.....

Your desperate claim that we cannot measure the temperature of stars
by analysing the absorption lines is not very convincing.

Emission lines might help....absorption lines provide willusory info about the
temperaturre of the absorbing layer.

...but obviously, the 'obvious' hasn't reached Norway yet.

And note one very important issue:
The temperature is derived from the absorption lines, and NOT
from the black body spectrum, which would have made my calculations
somewhat circular.

...that's not what I read.
Most of the lines are emission....lots of Fe, etc.,...
Anyway it probably makes little difference.

Your ignorance shows again, Henri.
All the 'spectral lines' in tab 4.1 are absorption lines.

What about the others?

and the radius curve in fig 3 in:
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf
Interferometry relies on all starlight moving at constant c wrt Earth.
Since that is not true, the technique can best be regarded as highly suspect.
The published cuvre is lousy anyway...nothing like a best fit....
Don't be such a blatant idiot, Henri.
If the BaTh predicts that it is possible to use a telescope
to measure the angular distance between two stars, then it also predicts
that it is possible to use an interferometer to measure the angular
diameter of a star. This is simple geometry, Henri. The geometry
is the same in both cases, and it doesn't matter what the speed
of light is.

Not so Paul. If the star is spinning..as it no doubt is..light from each side
will move at diffferent speeds towards earth..causing a willusory phase
difference.

This is mindless babble, Henri.
The "phase difference of light from different sides of the star"
is a meaningless concept - there is no such thing!
To compare the 'phase' of the light (noise!)
from two uncorrelated noise sources is meaningless!
What the hell should you compare?
And why do you think it is done anywhere?

Anyway, how can interferometry work at all?

It is indeed apparent that you don't understand how!

It requires coherent light and a split beam....or rather a split photon.....
Even you wouldn't claim that the same photon is emitted from both sides of the
star.

Should I laugh or should I cry?
I think I will laugh - a resigned laugh - this idiot is beyond reason.

Read my other posting.

I hope it's better than this one..



The two curves can be produced using BaTh. They differ only in eccentricity and
yaw angle..as well as phase.

So why did you call this explanation 'nonsensical'?

Because it is all based on willusroy data..

The fact that their
brigthness curves match those of stars in an elliptical orbit of e ~ 0.15-.25
and yaw angle -50-70 is purely coincidental.

Do you still find Planck's blackbody radiation law nonsensical, Henri?
Planck's law was empirically derived for what is assumed to be a perfect black
body.
Planck's law can be derived from the assumption (postulate) that
the energy of the oscillators in the radiating body is quantized.

Stars vary considerably and not many fit that curve well at all.
Nonsense.
All stellar spectra are black body spectra with absorption and
emission lines due to the fact that the black body radiation
from the stellar photosphere has to go through the higher layers
of the stellar atmosphere.

...Now even your colleagues will correct you on that one Paul..

Henri, zip your mouth, your ignorance shows.

The spectrum is given primarily by the surface temperature of
the star, and varies remarkably little between stars with equal
temperatures.

...but not variables....

- but not variables - what?

We can sum it up thus:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.

dream on....

..is what you say when you don't know what to say, right?

Whether you like it or not, these are indisputable FACTS:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.

IF.....??

The K and V curves are those of light emitted by a surface moving with the
radial velocities of Keplerian orbits. Both curves differ in 'yaw angle' and
'eccentricity', which in the case of a huffpuff, indicate the hysteresis in the
movement and the different characteristics between the in and out movements.
The ~80 degree phase difference shows that the two bands originate from
different layers at different times.

If you still haven't fathomed that, you better read again
my original posting where I prove the above to be facts.

Naturally the three willusions are consistent with each other since they are
caused by the same factors.

As the primary, measured data, I will use the temperature in fig. 4.3 in:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf
and the radius curve in fig 3 in:
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

yes....all wrong.....
.......very sad really to see so much effort wasted because of Einstein.

The question is:
What will the K (2.2u) and V (0.5u) light curves be
according to Planck's blackbody radiation law?

The result is shown in the table below. Here a
Int = the surface radiation intensity relative to the intensity at phase 0.
Lum = the luminosity (intensity*area) relative to the luminosity at phase 0.
Mag = the magnitude relative to the magnitude at phase 0.

The Intensity is calculated from Planck's black body radiation law.
Planck(T,lambda). (Look it up if you don't know it.)

It's in my physics book. Would you like a copy?

K Int = Planck(T,2.2u)/Planck(5600,2.2u)
V int = Planck(T,0.5u)/Planck(5600,0.5u)

Phase: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Temp: 5600 5550 5250 5050 4950 4900 4850 4950 5050 5400
Radius: 2.78 2.88 3.10 3.20 3.20 3.15 3.08 2.95 2.77 2.62
Area: 1.00 1.07 1.24 1.32 1.32 1.28 1.23 1.13 0.99 0.89
K Int: 1.00 0.98 0.89 0.84 0.81 0.79 0.78 0.81 0.84 0.94
K lum: 1.00 1.06 1.11 1.11 1.07 1.02 0.95 0.91 0.83 0.83
K mag: 0.00 -0.06 -0.12 -0.11 -0.07 -0.02 0.05 0.11 0.20 0.20
V Int: 1.00 0.95 0.71 0.57 0.51 0.48 0.45 0.51 0.57 0.83
V lum: 1.00 1.02 0.88 0.76 0.67 0.61 0.55 0.57 0.57 0.73
V mag: 0.00 -0.03 0.14 0.31 0.43 0.53 0.64 0.61 0.62 0.34

Compare K mag and V mag to the curves in fig.1 in
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The fit is so good that one could think I have cheated.

The fit is good because one set of the above figures was obtained form the
others.
All you have done is reversed the procedure..you have gone round in a
circle....

But I haven't. You can check the calculations yourself,
if you don't believe me.
The fit is good simply because Planck's black body radiation law
is correct, and a Cepheid is what it is known to be - a pulsating star.

Whether or not it is pulsating is not important. Its luminosity variation is
due primarily to cyclical changes in c+v.....as is the case for most variable
stars.


Paul



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

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.

On Wed, 05 Sep 2007 23:26:45 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Mon, 03 Sep 2007 22:37:52 +0200, "Paul B. Andersen"
wrote:


Imagine a 600m aperture reflector telescope.
This will have a resolution good enough to image the disk
of stars like l Carinae, and you can directly measure
the angular diameter of the star.

.....not through the atmosphere...

You are not very updated, Henri.
All modern large telescopes have adaptive optics compensating
for the phase shifts in the atmosphere.
Without adaptive optics, the resolution of Earth based telescopes
is limited to 0.5 - 1 arcsecs regardless of their size.
With adaptive optics, the resolution is very close to
the theoretical limit given by the size of the telescope.
The 10m Keck telescope has a resolution of 0.04 arcsecs in IR,
the 8.2m Subaru telescope has a resolution of 0.07 arcsecs in IR.
In both cases, this is very close to the theoretical limit.

The resolution of Earth based telescopes is now several times
better than the HST (because they are several times bigger.)

The VLTI (Very Large Telescope Interferomet) at
the European Southern Observatory has adaptive optics.

Well now you have to explain how a single photon can extend over 10 metres.

How does very low level starlight make a 'wavefront' at this distance?


How will the image now look?
It will still be basically the same image, and we can
still directly measure the angular diameter of the star.
The image will have fringes in it, though.
But there is no problem to measure the length of a zebra.

Well you raise an interesting point...how can both sides of a 600m telescope
pick up the same 'wavefront'?
If the wavefront includes of contributions from photons emitted from both sides
of a star, what made all the light coherent...is that part of the speed
unification process? I wouldn't be surprised....

You are indeed very confused, Henri. :-)
What you are saying above is that you don't understand how
a telescope (or your camera) can project an image at the CCD.

"If the wavefront hitting my camera includes light emitted from
both ears of my model, what made all the light coherent.." :-)

We are talking about interferometers remember.

1. The light from a star (or anything but a laser) isn't coherent,
it is noise, that is the amplitude and frequency vary arbitrarily.
But even such a wave have surfaces of equal phase, and the wave that
is emitted from a _point_ have a spherical surface of equal phase
(let's call it "wavefront"). The tiny fraction of the sphere that
hits our telescope will be a plane. And this plane wavefront will
be focused at a point on the CCD. Where the point will be is given
by the angle between the wavefront and the lens/mirror
2. Given that the 600m telescope is capable of resolving the star,
that is image it as a disc rather than a point, then a point
at one side of the star will be focused on one side of the image,
while a point on the other side of the star will be focused
on the other side of the image.
That is, the wavefront from a point on one side of the star
will have an angle compared to the wavefront from a point on
the other side of the star, so they will be focused at
different points on the CCD.
_That's how telescopes and cameras work._

Very good Paul. Nice to know they teach some optics in Norway.

But you must know this, Henri?

I DO Paul...and I also know that light across the aperture is virtually just
'noise'. It certainly isn't all in phase....If you think it is, please explain
the mechanism that brought it all into phase.

It doesn't answer the questions about interferometry.

Or is it no limit to your ignorance?
You discuss fiercely about how interferometers 'really' work
(dizzy photons), but have no clue about how a lens/mirror can
project an image on a screen(CCD)?
Amazing!

Paul, interferometers usually consist of at least a pair of receivers spaced as
far apart as practical.

If the angle subtended by a star is to be resolved by interference between
images from the two, the white light entering each must be coherent right
across the whole wavefront. How can that happen Paul?

I suppose you want to talk about one narrow spectral line rather than white
noise...that's OK. ...but I think interferometry uses radio rather than visible
light, CMIIW.

The BaTh predicts that interferometers like the above work,
because the speed of the light is utterly irrelevant.
Any theory that predicts that telescopes work, predicts
that interferometers work.

Just tell me how photons emitted from oposite sides of a star can end up in
phase over a 600m wavefrant...

This stupidity again!
Why the hell would you like the light from both sides of the star
or from both ears of your model 'to end up in phase' over the aperture
of the telescope or your camera?

How confused can you get?

Paul, you are really lost.
Interferometry requires TWO receivers.

So we have indeed quite directly measured the diameters
of Cepheids and Miras, and literally seen that they are pulsating.
Any talk about dizzy photons, willusions or other stupidities
can't change this fact.

.....dream on Paul....

Henri, you will have to accept that telescopes and cameras work.
No idiotic babble about 'light emitted from different parts of
the object ending up in phase over the aperture of the telescope'
can change the fact that TELESCOPES WORK!

And a 600m telescope would be able to image l Carinae as a disc.

....but we don't have such telescopes and it doesn't happen.

With this fact in mind, read the following again:

Now we cut out two 8m diameter circular disks at opposite
sides of the rim of our giant telescope mirror.
We keep these disks, and remove the rests of the giant mirror.
So we have two 8m mirrors separated by 584 m.
They are still focusing at the same spot.

How will the image now look?
It will still be basically the same image, and we can
still directly measure the angular diameter of the star.
The image will have fringes in it, though.

I'll bet it will...Fringes caused by different light speeds...

Note this, Henri.
The image from two 8m mirrors 600m apart is the same
as the image from a 600m mirror, but for a few fringes
in the image! You can directly measure the diameter
of the star by measuring the diameter of image!

No you can't. It's still far too small.

To refute this is idiocy. Several of these instruments
are now in daily use. To claim that they don't work
is as idiotic as claiming that cameras don't work.

But if your religion demands it, you will deny anything.
Right?

Paul, it matters not one iota whether or not the star goes huff puff.
The main cause of the luminosity variation is cyclic c+.
....or do you still believe that most star curves can be matched with the BaTh
out of pure coincidence?

Paul



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

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.

Henri Wilson wrote:
On Thu, 06 Sep 2007 00:10:30 +0200, "Paul B. Andersen"
wrote:
Whether you like it or not, these are indisputable FACTS:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.

IF.....??

The K and V curves are those of light emitted by a surface moving with the
radial velocities of Keplerian orbits. Both curves differ in 'yaw angle' and
'eccentricity', which in the case of a huffpuff, indicate the hysteresis in the
movement and the different characteristics between the in and out movements.
The ~80 degree phase difference shows that the two bands originate from
different layers at different times.

You are trying to make the circularity you are accusing me of,
namely starting with the K- and V-light curves, and trying to
guess what the parameters in your model must be to predict
the observed observed data.

But despite of this circularity, you do not succeed.
There simply IS no way your model can produce the observed
data.


If you still haven't fathomed that, you better read again
my original posting where I prove the above to be facts.

Naturally the three willusions are consistent with each other since they are
caused by the same factors.

As the primary, measured data, I will use the temperature in fig. 4.3 in:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf

Note that the temperature curve is inferred by
the absorption lines - and defenitely NOT by
the measurements

and the radius curve in fig 3 in:
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

yes....all wrong.....
......very sad really to see so much effort wasted because of Einstein.

Note that radius curve is confirmed by spectroscopic
measurement, which is a very direct method of measuring
the diameter.

This means that the input data do not depend on
measurements of the black body spectrum!

The question is:
What will the K (2.2u) and V (0.5u) light curves be
according to Planck's blackbody radiation law?

The result is shown in the table below. Here a
Int = the surface radiation intensity relative to the intensity at phase 0.
Lum = the luminosity (intensity*area) relative to the luminosity at phase 0.
Mag = the magnitude relative to the magnitude at phase 0.

The Intensity is calculated from Planck's black body radiation law.
Planck(T,lambda). (Look it up if you don't know it.)

It's in my physics book. Would you like a copy?

K Int = Planck(T,2.2u)/Planck(5600,2.2u)
V int = Planck(T,0.5u)/Planck(5600,0.5u)

Phase: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Temp: 5600 5550 5250 5050 4950 4900 4850 4950 5050 5400
Radius: 2.78 2.88 3.10 3.20 3.20 3.15 3.08 2.95 2.77 2.62
Area: 1.00 1.07 1.24 1.32 1.32 1.28 1.23 1.13 0.99 0.89
K Int: 1.00 0.98 0.89 0.84 0.81 0.79 0.78 0.81 0.84 0.94
K lum: 1.00 1.06 1.11 1.11 1.07 1.02 0.95 0.91 0.83 0.83
K mag: 0.00 -0.06 -0.12 -0.11 -0.07 -0.02 0.05 0.11 0.20 0.20
V Int: 1.00 0.95 0.71 0.57 0.51 0.48 0.45 0.51 0.57 0.83
V lum: 1.00 1.02 0.88 0.76 0.67 0.61 0.55 0.57 0.57 0.73
V mag: 0.00 -0.03 0.14 0.31 0.43 0.53 0.64 0.61 0.62 0.34

Compare K mag and V mag to the curves in fig.1 in
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The fit is so good that one could think I have cheated.

The fit is good because one set of the above figures was obtained form the
others.
All you have done is reversed the procedure..you have gone round in a
circle....

Neither the temperature curve nor the radius curve
are inferred from the K- and V-light curves.
There is no circularity.

But I haven't. You can check the calculations yourself,
if you don't believe me.
The fit is good simply because Planck's black body radiation law
is correct, and a Cepheid is what it is known to be - a pulsating star.

Whether or not it is pulsating is not important. Its luminosity variation is
due primarily to cyclical changes in c+v.....as is the case for most variable
stars.

This is an indisptable FACT, Henri:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.


I don't believe in coincidences.
So when
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.
it is because
l Carinae is a pulsating star with temperature curve and
diameter curve as measured.

Paul

Henri Wilson wrote:
On Wed, 05 Sep 2007 23:26:45 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Mon, 03 Sep 2007 22:37:52 +0200, "Paul B. Andersen"
wrote:


Imagine a 600m aperture reflector telescope.
This will have a resolution good enough to image the disk
of stars like l Carinae, and you can directly measure
the angular diameter of the star.
.....not through the atmosphere...
You are not very updated, Henri.
All modern large telescopes have adaptive optics compensating
for the phase shifts in the atmosphere.
Without adaptive optics, the resolution of Earth based telescopes
is limited to 0.5 - 1 arcsecs regardless of their size.
With adaptive optics, the resolution is very close to
the theoretical limit given by the size of the telescope.
The 10m Keck telescope has a resolution of 0.04 arcsecs in IR,
the 8.2m Subaru telescope has a resolution of 0.07 arcsecs in IR.
In both cases, this is very close to the theoretical limit.

The resolution of Earth based telescopes is now several times
better than the HST (because they are several times bigger.)

The VLTI (Very Large Telescope Interferomet) at
the European Southern Observatory has adaptive optics.

Well now you have to explain how a single photon can extend over 10 metres.

How does very low level starlight make a 'wavefront' at this distance?

The wave-particle duality is an experimentally proven fact.
Whether I, you or anybody else can 'explain' it is irrelevant.
Keck works, and can collect photons from distant sources
literally one by one, and each an every photon emitted from
the same point source is focused on the same small spot which
only can be explained by the 10m aperture.

How will the image now look?
It will still be basically the same image, and we can
still directly measure the angular diameter of the star.
The image will have fringes in it, though.
But there is no problem to measure the length of a zebra.
Well you raise an interesting point...how can both sides of a 600m telescope
pick up the same 'wavefront'?
If the wavefront includes of contributions from photons emitted from both sides
of a star, what made all the light coherent...is that part of the speed
unification process? I wouldn't be surprised....
You are indeed very confused, Henri. :-)
What you are saying above is that you don't understand how
a telescope (or your camera) can project an image at the CCD.

"If the wavefront hitting my camera includes light emitted from
both ears of my model, what made all the light coherent.." :-)

We are talking about interferometers remember.

1. The light from a star (or anything but a laser) isn't coherent,
it is noise, that is the amplitude and frequency vary arbitrarily.
But even such a wave have surfaces of equal phase, and the wave that
is emitted from a _point_ have a spherical surface of equal phase
(let's call it "wavefront"). The tiny fraction of the sphere that
hits our telescope will be a plane. And this plane wavefront will
be focused at a point on the CCD. Where the point will be is given
by the angle between the wavefront and the lens/mirror
2. Given that the 600m telescope is capable of resolving the star,
that is image it as a disc rather than a point, then a point
at one side of the star will be focused on one side of the image,
while a point on the other side of the star will be focused
on the other side of the image.
That is, the wavefront from a point on one side of the star
will have an angle compared to the wavefront from a point on
the other side of the star, so they will be focused at
different points on the CCD.
_That's how telescopes and cameras work._

Very good Paul. Nice to know they teach some optics in Norway.

But you must know this, Henri?

I DO Paul...and I also know that light across the aperture is virtually just
'noise'. It certainly isn't all in phase....If you think it is, please explain
the mechanism that brought it all into phase.

Of bloody course it isn't "all in phase". If it were, all the
light would be focused in a single point! Telescopes make _images_!
So why the hell are you stating such an irrelevant triviality?

It doesn't answer the questions about interferometry.

See below.

Or is it no limit to your ignorance?
You discuss fiercely about how interferometers 'really' work
(dizzy photons), but have no clue about how a lens/mirror can
project an image on a screen(CCD)?
Amazing!

Paul, interferometers usually consist of at least a pair of receivers spaced as
far apart as practical.
If the angle subtended by a star is to be resolved by interference between
images from the two, the white light entering each must be coherent right
across the whole wavefront. How can that happen Paul?

You demonstrate yet again that you don't know how an interferometer works.

See below.

I suppose you want to talk about one narrow spectral line rather than white
noise...that's OK. ...but I think interferometry uses radio rather than visible
light, CMIIW.

Henri, the interferometer that started this discussion was the interferometer
which measured the diameter of l Carinae. It is the VLTI
(Very Large Telescope Interferometer) at the European Southern Observatory.
This is the kind of interferometer I am talking about.

The BaTh predicts that interferometers like the above work,
because the speed of the light is utterly irrelevant.
Any theory that predicts that telescopes work, predicts
that interferometers work.
Just tell me how photons emitted from oposite sides of a star can end up in
phase over a 600m wavefrant...
This stupidity again!
Why the hell would you like the light from both sides of the star
or from both ears of your model 'to end up in phase' over the aperture
of the telescope or your camera?

How confused can you get?

Paul, you are really lost.
Interferometry requires TWO receivers.

Indeed.
That doesn't make your babble above less idiotic.
See below.

So we have indeed quite directly measured the diameters
of Cepheids and Miras, and literally seen that they are pulsating.
Any talk about dizzy photons, willusions or other stupidities
can't change this fact.
.....dream on Paul....
Henri, you will have to accept that telescopes and cameras work.
No idiotic babble about 'light emitted from different parts of
the object ending up in phase over the aperture of the telescope'
can change the fact that TELESCOPES WORK!

And a 600m telescope would be able to image l Carinae as a disc.

...but we don't have such telescopes and it doesn't happen.

Are you going to insist that a 600m telescope can't work? :-)
Of course you don't.
So read on.

With this fact in mind, read the following again:

Now we cut out two 8m diameter circular disks at opposite
sides of the rim of our giant telescope mirror.
We keep these disks, and remove the rests of the giant mirror.
So we have two 8m mirrors separated by 584 m.
They are still focusing at the same spot.

How will the image now look?
It will still be basically the same image, and we can
still directly measure the angular diameter of the star.
The image will have fringes in it, though.

I'll bet it will...Fringes caused by different light speeds...

Note this, Henri.
The image from two 8m mirrors 600m apart is the same
as the image from a 600m mirror, but for a few fringes
in the image! You can directly measure the diameter
of the star by measuring the diameter of image!

No you can't. It's still far too small.

The resolution of a 600m telescope is ca. 0.5 mas.
The diameter of l Carinae is ca. 3 mas.
mas = milli-arc-sec

It is simply a fact that interferometers like the VLTI do
image stars like l Carinae as a disk which you can measure
the diameter of.

To refute this is idiocy. Several of these instruments
are now in daily use. To claim that they don't work
is as idiotic as claiming that cameras don't work.

But if your religion demands it, you will deny anything.
Right?

Paul, it matters not one iota whether or not the star goes huff puff.
The main cause of the luminosity variation is cyclic c+.
...or do you still believe that most star curves can be matched with the BaTh
out of pure coincidence?

The issue is your claim that the diameter of stars like l Carinae
cannot be measured by interferometric measurements.

Do you now understand that it can?

If telescopes work, so do interferometers.
The principle is the very same.
Have you now fathomed that, or will you still state stupidities like:
"Just tell me how photons emitted from opposite sides of a star can end up in
phase over a 600m wavefront".

Paul

On Sun, 09 Sep 2007 21:39:34 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 05 Sep 2007 23:26:45 +0200, "Paul B. Andersen"
wrote:

The resolution of Earth based telescopes is now several times
better than the HST (because they are several times bigger.)

The VLTI (Very Large Telescope Interferomet) at
the European Southern Observatory has adaptive optics.

Well now you have to explain how a single photon can extend over 10 metres.

How does very low level starlight make a 'wavefront' at this distance?

The wave-particle duality is an experimentally proven fact.
Whether I, you or anybody else can 'explain' it is irrelevant.
Keck works, and can collect photons from distant sources
literally one by one, and each an every photon emitted from
the same point source is focused on the same small spot which
only can be explained by the 10m aperture.

Pathetic!
In future, if you can't answer, please don't try ...



We are talking about interferometers remember.

1. The light from a star (or anything but a laser) isn't coherent,
it is noise, that is the amplitude and frequency vary arbitrarily.
But even such a wave have surfaces of equal phase, and the wave that
is emitted from a _point_ have a spherical surface of equal phase
(let's call it "wavefront"). The tiny fraction of the sphere that
hits our telescope will be a plane. And this plane wavefront will
be focused at a point on the CCD. Where the point will be is given
by the angle between the wavefront and the lens/mirror
2. Given that the 600m telescope is capable of resolving the star,
that is image it as a disc rather than a point, then a point
at one side of the star will be focused on one side of the image,
while a point on the other side of the star will be focused
on the other side of the image.
That is, the wavefront from a point on one side of the star
will have an angle compared to the wavefront from a point on
the other side of the star, so they will be focused at
different points on the CCD.
_That's how telescopes and cameras work._

Very good Paul. Nice to know they teach some optics in Norway.

But you must know this, Henri?

I DO Paul...and I also know that light across the aperture is virtually just
'noise'. It certainly isn't all in phase....If you think it is, please explain
the mechanism that brought it all into phase.

Of bloody course it isn't "all in phase". If it were, all the
light would be focused in a single point! Telescopes make _images_!
So why the hell are you stating such an irrelevant triviality?

I'm trying to stimulate your brain into some kind of action....

It doesn't answer the questions about interferometry.

See below.

I can hardly wait....

Or is it no limit to your ignorance?
You discuss fiercely about how interferometers 'really' work
(dizzy photons), but have no clue about how a lens/mirror can
project an image on a screen(CCD)?
Amazing!

Paul, interferometers usually consist of at least a pair of receivers spaced as
far apart as practical.
If the angle subtended by a star is to be resolved by interference between
images from the two, the white light entering each must be coherent right
across the whole wavefront. How can that happen Paul?

You demonstrate yet again that you don't know how an interferometer works.

See below.

I can hardly wait....

I suppose you want to talk about one narrow spectral line rather than white
noise...that's OK. ...but I think interferometry uses radio rather than visible
light, CMIIW.

Henri, the interferometer that started this discussion was the interferometer
which measured the diameter of l Carinae. It is the VLTI
(Very Large Telescope Interferometer) at the European Southern Observatory.
This is the kind of interferometer I am talking about.

OK.
It detects something.
I say it detects changes in light speed along with other willusions.

The BaTh predicts that interferometers like the above work,
because the speed of the light is utterly irrelevant.
Any theory that predicts that telescopes work, predicts
that interferometers work.
Just tell me how photons emitted from oposite sides of a star can end up in
phase over a 600m wavefrant...
This stupidity again!
Why the hell would you like the light from both sides of the star
or from both ears of your model 'to end up in phase' over the aperture
of the telescope or your camera?

How confused can you get?

Paul, you are really lost.
Interferometry requires TWO receivers.

Indeed.
That doesn't make your babble above less idiotic.
See below.

I can hardly wait....

So we have indeed quite directly measured the diameters
of Cepheids and Miras, and literally seen that they are pulsating.
Any talk about dizzy photons, willusions or other stupidities
can't change this fact.
.....dream on Paul....
Henri, you will have to accept that telescopes and cameras work.
No idiotic babble about 'light emitted from different parts of
the object ending up in phase over the aperture of the telescope'
can change the fact that TELESCOPES WORK!

And a 600m telescope would be able to image l Carinae as a disc.

...but we don't have such telescopes and it doesn't happen.

Are you going to insist that a 600m telescope can't work? :-)
Of course you don't.
So read on.

I can hardly wait....

With this fact in mind, read the following again:

Now we cut out two 8m diameter circular disks at opposite
sides of the rim of our giant telescope mirror.
We keep these disks, and remove the rests of the giant mirror.
So we have two 8m mirrors separated by 584 m.
They are still focusing at the same spot.

How will the image now look?
It will still be basically the same image, and we can
still directly measure the angular diameter of the star.
The image will have fringes in it, though.

I'll bet it will...Fringes caused by different light speeds...

Note this, Henri.
The image from two 8m mirrors 600m apart is the same
as the image from a 600m mirror, but for a few fringes
in the image! You can directly measure the diameter
of the star by measuring the diameter of image!

No you can't. It's still far too small.

The resolution of a 600m telescope is ca. 0.5 mas.
The diameter of l Carinae is ca. 3 mas.
mas = milli-arc-sec

It is simply a fact that interferometers like the VLTI do
image stars like l Carinae as a disk which you can measure
the diameter of.

You still haven't explained how the weak starlight that enters both detectors
is coherent. Since it comes from many parts of the star, I don't see that it
can be .....UNLESS of course you accept my unification theory.

To refute this is idiocy. Several of these instruments
are now in daily use. To claim that they don't work
is as idiotic as claiming that cameras don't work.

But if your religion demands it, you will deny anything.
Right?

Paul, it matters not one iota whether or not the star goes huff puff.
The main cause of the luminosity variation is cyclic c+.
...or do you still believe that most star curves can be matched with the BaTh
out of pure coincidence?

The issue is your claim that the diameter of stars like l Carinae
cannot be measured by interferometric measurements.

Do you now understand that it can?

I certainly do understand.... and the answer is clearly NO, IT CANNOT..

If telescopes work, so do interferometers.
The principle is the very same.

This reveals your ignorance of optics..

Have you now fathomed that, or will you still state stupidities like:
"Just tell me how photons emitted from opposite sides of a star can end up in
phase over a 600m wavefront".

Interferometry requires coherent light.
Please provide an explanation as to how light emitted from many parts of a
star's surface, maybe at slightly different times, can end up in phase 1800LYs
away and over a distance of 600m.

Paul



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

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.

On Sun, 09 Sep 2007 19:00:27 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Thu, 06 Sep 2007 00:10:30 +0200, "Paul B. Andersen"
wrote:
Whether you like it or not, these are indisputable FACTS:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.

IF.....??

The K and V curves are those of light emitted by a surface moving with the
radial velocities of Keplerian orbits. Both curves differ in 'yaw angle' and
'eccentricity', which in the case of a huffpuff, indicate the hysteresis in the
movement and the different characteristics between the in and out movements.
The ~80 degree phase difference shows that the two bands originate from
different layers at different times.

You are trying to make the circularity you are accusing me of,
namely starting with the K- and V-light curves, and trying to
guess what the parameters in your model must be to predict
the observed observed data.

But despite of this circularity, you do not succeed.
There simply IS no way your model can produce the observed
data.

I have already produced the curves and the REAL data required to produce them.

If you still haven't fathomed that, you better read again
my original posting where I prove the above to be facts.

Naturally the three willusions are consistent with each other since they are
caused by the same factors.

As the primary, measured data, I will use the temperature in fig. 4.3 in:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf

Note that the temperature curve is inferred by
the absorption lines - and defenitely NOT by
the measurements

....what measurements?

and the radius curve in fig 3 in:
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

yes....all wrong.....
......very sad really to see so much effort wasted because of Einstein.

Note that radius curve is confirmed by spectroscopic
measurement, which is a very direct method of measuring
the diameter.



This means that the input data do not depend on
measurements of the black body spectrum!

It depends on constant light speed.

The question is:
What will the K (2.2u) and V (0.5u) light curves be
according to Planck's blackbody radiation law?

The result is shown in the table below. Here a
Int = the surface radiation intensity relative to the intensity at phase 0.
Lum = the luminosity (intensity*area) relative to the luminosity at phase 0.
Mag = the magnitude relative to the magnitude at phase 0.

The Intensity is calculated from Planck's black body radiation law.
Planck(T,lambda). (Look it up if you don't know it.)

It's in my physics book. Would you like a copy?

K Int = Planck(T,2.2u)/Planck(5600,2.2u)
V int = Planck(T,0.5u)/Planck(5600,0.5u)

Phase: 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Temp: 5600 5550 5250 5050 4950 4900 4850 4950 5050 5400
Radius: 2.78 2.88 3.10 3.20 3.20 3.15 3.08 2.95 2.77 2.62
Area: 1.00 1.07 1.24 1.32 1.32 1.28 1.23 1.13 0.99 0.89
K Int: 1.00 0.98 0.89 0.84 0.81 0.79 0.78 0.81 0.84 0.94
K lum: 1.00 1.06 1.11 1.11 1.07 1.02 0.95 0.91 0.83 0.83
K mag: 0.00 -0.06 -0.12 -0.11 -0.07 -0.02 0.05 0.11 0.20 0.20
V Int: 1.00 0.95 0.71 0.57 0.51 0.48 0.45 0.51 0.57 0.83
V lum: 1.00 1.02 0.88 0.76 0.67 0.61 0.55 0.57 0.57 0.73
V mag: 0.00 -0.03 0.14 0.31 0.43 0.53 0.64 0.61 0.62 0.34

Compare K mag and V mag to the curves in fig.1 in
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf


The fit is good simply because Planck's black body radiation law
is correct, and a Cepheid is what it is known to be - a pulsating star.

Whether or not it is pulsating is not important. Its luminosity variation is
due primarily to cyclical changes in c+v.....as is the case for most variable
stars.

This is an indisptable FACT, Henri:
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.

Paul, Planck's Law says nothing about variations in a star's luminosity.

I don't believe in coincidences.
So when
The K and V light curves from l Carinae are exactly as
Planck's law predicts them to be if the star is
a pulsating star with temperature curve and diameter curve
as measured.
it is because
l Carinae is a pulsating star with temperature curve and
diameter curve as measured.

Even if L Car does happen to pulsate a little..and that's debatable...most of
its lumnosity variation is due to ADoppler..derived from variable light speed.
This is proved by the fact that I can match the K and V curves exactly using
nothing but BaTh.

Paul



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

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.

"Henri Wilson" HW@.... wrote in message
...
On Thu, 30 Aug 2007 10:31:32 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
news
On Wed, 29 Aug 2007 18:44:27 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
m...
On Sat, 25 Aug 2007 13:32:44 +0100, "George Dishman"

For a start, I certainly DON'T accept that the radius varies by 12%

Tough, that is the value directly measured by ESO.

it is wrong.

Rubbish, the speed of light in air is the same
as in conventional theory so the results are
the same.

Light traves a long way before it reaches Earth's air.

But that has no effect on the interferometer,
all of it is on Earth ;-)

George, presumably the interference is caused by the angle subtended by
the
star.

Nope. You have covered some of this with paul but
you have only partly grasped the situation. Look
again at the setup:

http://tinyurl.com/3dybf3

and compare it with this

http://www.georgedishman.f2s.com/Hen...ic_grating.gif

As with any interferometer the pattern depends on
the distance between the paths at the receiving
end. A maximum occurs where the path length
difference is a multiple of a wavelength. The same
is true here but instead of simple rulings on a
grating you have two separate telescopes, ANTU and
MELIPAL, providing the paths.

Remember when we talked of the grating, I made the
point that a single photon would be deflected by an
angle that depended on its frequency or wavelength
adn a distribution plot of where photons land gives
the usual pattern. The same is true again, light from
one side of the star passes through both telescopes
and produces a set of fringes. Completely independently
light from the other side also produces a set of
fringes but because the source is slightly displaced,
so are the fringes. As a result, the minima don't
occur at exactly the same place so they don't go to
zero. The contrast ratio then gives an indication of
the displacement as a fraction of a fringe and hence
of the angular width of the star.

If it is rotating, the lght from both sides will be phase shifted as they
arrive due to c+/-v.

The light arrives at c/n where n is the refractive
index of the air around the telescopes. The phase
difference across the system (i.e. between the two
telescopes) depends on their separation and that
speed.

I doubt if anything would show up at 50000 LYs...

Probably not, that's why they started with L Car,
it is one of the closest and largest Cepheids and
subtends the largest angle.

Interferomery will give a distorted answer.

Nope, there is no distortion introduced
by ballistic theory.

I think it is fair to assume all stars are rotatiing.

Sure, but photons from one side of the star arrive
at some speed and get deflected through some angle
by the interferometer. What speed it left the star
makes no difference to the pattern. The same is true
for photons from the other side, every photon acts
independently.

Sorry Henry, you have to do better than hand waving.
The speed at the interferometer is the same across
the instrument so the interference pattern is
unaffected.

George, the technique is highly suspect at best.

Garbage, it is no more suspect than a grating.

Add variable light speed and
it becomes almost useless.

It has no effect, you only want to wave it
away because you cannot stomach the truth.

the star's rotation stuffs up the whole process.

Not in the slightest.

there would even be a willusory temperature variation due to ADopler
shifting
of the Planck curve..

Nope, the shift is only 0.01%. The K band is from
2200nm to 2400nm so the median shift is 0.22nm.
How much does that change the intensity in the band
for a Planck curve at ~6000K? It is utterly negligible.

Don't be so hasty George.
The Planck curve deals with PHOTON DENSITY in a particular band.

Intensity Henry.

Same thing.

Nope, your ignorance is showing again.

Photon density variation due to ADoppler DOES NOT include my 'K'
factor...so
your figure of 0.01 is not anywhere near the correct one.

There you go again! You are completely ignoring the willusion....

Check the attribution, you just disagreed
with yourself.

I am saying
Cepheid surface speeds are typically less than 30km/s
so 0.01% is an upper limit. Whether that is caused by
VDoppler or ADoppler doesn't matter, the shift is no
more than that value. That means no more than 0.24nm
worth of the band moves out at one end while about the
same amount moves in at the other.

George, you will never learn anything about cepheids from willusory data..

If it is shifted by 0.01%, that's how much
falls off one end of the filter and into the
other.

This is going to become pretty complicated so I will think about it.

Do that, you are obviously missing the point at the
moment.

you are mssing the willusions...

Nope, 0.01% is the shift regardless of cause,
think about it.

George, quite clearly, if L Car is a huffpuff, its maximum temperture
should
occur about 30 degrees BEFORE minimum radius....when the 'exploding'
core
bangs

When I
match KNOWN curves, you say I just fiddle with a curve matching program
till I
get the right answer.
You should be consistent George.

I am. You know that for simultaneous equations
you can find a solution if you have as many
equations ars you have variables. You have
numerous parameters you can alter to get a fit
and basically if you have say ten variable, you
can do a Fourier fit of up to the fifth harmnonic
with sin and cosine terms (or amplitude and phase)
for each.

You don't understand how narrow are the program's conditions.

'Close Encounters' Henry ;-)

For Cepheid models you have basically the mass
of the star and to a degree the elemental
abundance. For any particular star you also have
the age but the model has to fit over the full
evolution of the star so that isn't really free
from a modelling point of view. Also mass, age
and chemistry can all be constrained by observation
so there is no significant scope for fiddling.

All the bserved data is willusory and cannot be assumed correct.

Wrong again, temperatures and subtended angle are
valid as I have explained to you several times.

...and have been pointing out that the velocity curve should be similar
in
shape an phase to the luminosity curve...but you never listen...

No, check the top of this post, you were arguing
that the luminosity peaked with the acceleration,
not the velocity.

That's correct

Well make your mind up.

That is why I keep telling you that the only
way you can get a valid analysis is to fit
your predicted curve for the observed velocity
and then work back to get the true velocity.

You still don't understand that I feed into the program the TRUE
orbital
parameters including velocity.

I know that. What you need to do is alter your
program so that it predicts what would be the
OBSERVED velocity curve based on spectral line
shift using the values you feed in and ballistic

Since I can produce the exact curves without including a temperature or
radius
changes, my conclusion could easily and quite justifiably be that
neither
changes occur...except in the minds of relativists.

Since both changes are directly observed, your
conclusion is wrong.


No George, all that is observed is a number of willusions.

Think about it for a change instead of repeating
your pointless mantra.

To justify it, you would
need to write down the equations and then solve
them to show that a star whoae temperature didn't
vary would produce a Planck-shaped curve over
multiple bands which varied _as_if_ the temperature
were changing due to some ballistic effect. You
can't do that because photon bunching due to
ADoppler and VDoppler is frequency independent.

That means we know the temperature _does_ change
and with it surface brightness so until you
subtract that part from the luminosity curve,
your results are badly flawed.

It seems you will never understand...maybe your holiday will help...

Nothing has changed, you are still making
content-free statements and not addressing
the actual situation.

From that you can integrate to get the true
radius or differentiate to get the true
acceleration and from those AND the temperature
AND the filter bandwidths you could then predict
the luminosity curves.

No George. You have it all back to front. I can calculate K for a star
by
comparing the ADoppler produced luminosity variation with he OBSERVED
fractional velocity change.

Not until you correct the luminosity for
temperature and radius effects.

These are the willusory temperature changes of course....

The temperature _value_ is valid, only the arrival
time (orbital phase) would be offset by the c+v
effect.

I cnt see our sun fluctuating in brigtness or radius....yet it would be
classed
as a variable by a distant observer.

It would appear to vary in luminosity but not in
radius or temperature which is what we are talking
about, try to keep a grasp of the conversation Henry.

....
Emission times cannot matter because the light
is uncorrelated anyway. Speed differences could
matter but the light is moving at the same c/n
value when it reaches the interferometer so there
is no real scope for a distortion that way that
I can see.

There will be a phase difference caused by different source speeds. If the
star
is a binary and the surrounding 'EM sphere' is steady, there will be
wavelength
differences between light coming from each side.

You might want to consider the overall setup:

http://tinyurl.com/3dybf3

No, it wont work..

But it does work Henry, they get fringes exactly
as all the theories say they will.

The are two separate processes. There is an acoustic pressure wave that
causes
adiabatic compression and temperature rise. As radius increases, there
is
also
an expansion that results in an adiabatic temperature DECREASE.

There is also simple heating due to the added
energy which is the more significant contribution.

That would require turbulent diffusion because thermal conductivty of
gasses is
quite small. Such diffusion would be far too slow.

The transfer is principally radiative but it is not
fast due to the opacity.

Frankly I cannot see any obvous connection between the acoustic wave and
your
supposed largish radius change...or surface temperature.

I'll try to find the eigenstate plots which
make it clear.

It would be more like a mass movement than an acoustic wave.....if it
occurred
at all.

Yes, that's what Jerry and I were telling you months
ago.

....
I don't see how a single photon could be emitted by both sides of a
star.
if it was, it would create NO interference.

No, no Henry each photon is emitted by a single
charged particle. Each photon passes through
both telescopes of the interferometer and
lands with a probability that depends on the
path length difference to create an interference
pattern matching the probability of landing at
some point. It is similar to the usual grating
equation. It is the overlaying of those patterns
from different parts of the star that alters the
contrast ratio of the fringes and tells us the
diameter.

It ignores the different c+v from both sides.

A single photon doesn't come from "both sides" and for
each photon it is only the speed at the interferometer
together with the frequency that determines the
wavelength, lambda_r:

http://www.georgedishman.f2s.com/Hen...ic_grating.gif

George

"Henri Wilson" HW@.... wrote in message
...
On Sun, 09 Sep 2007 21:39:34 +0200, "Paul B. Andersen"
wrote:
Henri Wilson wrote:
On Wed, 05 Sep 2007 23:26:45 +0200, "Paul B. Andersen"
wrote:

The resolution of Earth based telescopes is now several times
better than the HST (because they are several times bigger.)

The VLTI (Very Large Telescope Interferomet) at
the European Southern Observatory has adaptive optics.

Well now you have to explain how a single photon can extend over 10
metres.

How does very low level starlight make a 'wavefront' at this distance?

The wave-particle duality is an experimentally proven fact.
Whether I, you or anybody else can 'explain' it is irrelevant.
Keck works, and can collect photons from distant sources
literally one by one, and each an every photon emitted from
the same point source is focused on the same small spot which
only can be explained by the 10m aperture.

Pathetic!
In future, if you can't answer, please don't try ...

....

You still haven't explained how the weak starlight that enters both
detectors
is coherent. Since it comes from many parts of the star,

He did, though perhaps he assumed too much of you, it
is coherent because each detection is of a _single_
photon and any photon is of course coherent with itself.

Have you now fathomed that, or will you still state stupidities like:
"Just tell me how photons emitted from opposite sides of a star can end up
in
phase over a 600m wavefront".

Interferometry requires coherent light.

Interferometry works with individual photons even
in this configuration:

http://tinyurl.com/3dybf3

George

"Henri Wilson" HW@.... wrote in message
...
On Wed, 29 Aug 2007 22:39:15 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
....
It is virtually the same as the luminosity curve.... upside down...

in: http://www.georgedishman.f2s.com/Henri/Cepheid_typ.png

the Lum varies by about 2.5:1 whilst the velocity varies by 1.3E-4:1,
making K = 5E-5

No comment George?
Don't understand maths again?

Pointless, the 2.5:1 luminosity variation is
dominated by temperature and radius changes
and you have to remove those before attempting
to work out K. Also, before you can work it out,
you need to say where it goes in the luminosity
equation and then solve for K.

I will calculate K for more stars in future to see if there is any
consistency.

Don't waste your time tossing meaningles numbers
around, you cannot calculate 'K' until you decide
where it goes in the equations.

You are yet to acknowledge that the Time for light to go from A to B is
not
constant.

You forget I am the one who has consistently corrected
your error of ignoring refractive index.

George