Why are the 'Fixed Stars' so FIXED?

"Henri Wilson" HW@.... wrote in message
news
On Thu, 30 Aug 2007 17:17:24 +0100, "George Dishman"

wrote:


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

The temperature of all layers should increase as the star contracts
under
gravity.

Sure, both factors operate.

I think you haven't previously looked at a
typical temperature curve.

I have....and it is willusory anyway...

Nope, other than time of arrival, the
temperature is a ratio of bands so isn't
affected.

The 'ratio of bands' is very sensitive to the type of radiator. Any
variation
from black body could have a profound effect.

Indeed and care must be taken for that reason
especially with local factors like absorption
by water and oxygen in the K band. These
effects are well known though, nobody ignores
them.

....but everybody seems to ignore the most critical factor..that of
variable
light speed....

People take the frequency dependence of the
refractive index into account where it has
an effect, e.g. in pulsar dispersion.

The cause doesn't matter, the shift is less than
0.01% or 0.22nm for K band when the filter is
400nm wide - completely negligible.

You cannot assume a consant emissivity for the changing surface layer
either.

The emissivity is 100% at the bottom of the
layer Henry, Kirchoff's law requires that.

Not if its temperature is continually changing.

Yes Henry, Kirchoff's law requires it.

Nor can you have photons with negative lengths...

That's your 'wave equation' of course....

No that's YOUR 'wave model', you have no wave equation.

Photons are particles.

Right, but you don't have an equation for that either.

No you aren't. You didn't even consider the main factor, the temperature
gradient in the water and its affect on viscosity....
We know the ball's volume will decrease nonlinearly and we can assume it
remains in temperature equilibrium with the water.

The sea's temperature changes only slightly with
depth after the first few tens of metres, and the
effect on the ball will be minimal. Viscosity
has no effect at all on the volume of the ball.

I know that George.

Thank goodness.

I'm talking about the rate of fall, ...

No, you were talking about the ball being compressed
by pressure as an analogy for a photon being squeezed
by the differential velocity due to acceleration at
the time of emission, or some such rubbish, you never
mentioned rate of fall.

I don't see how Kirchoff's law really comes into this. Sure the
emissivity
of
the surface is likely to change with both temperature and density but
the
law
will still hold.

Since the gas is a black body radiator, it must
also be a perfect absorber. As the density rises,
it becomes completely opaque which is why you
cannot see through to a second layer.

they are big assumptions...

Nope, they are results confirmed by lab tests.

The models at first could not get the 10 day
period right for the in-phase 'bump' no matter
how people tried to adjust them. The opacity
of He++ was rechecked and found to be wrong
and that solved the problem. The essence of a
good model is that is _cannot_ be made to match
unless the parameters are valid, unlike your
excellent match to the theme from Close
Encounters with your "Keplerian Orbits Only"
program.

......so you believe that cepheid curves are Keplarian out of pure
coincidence?

No, I believe you have added so many adjustable
parameters in your program that you can fit any
curve, Keplerian or not.

George, the well known cepheid curve is Keplerian...whether you like it or
not...

Nope, but your program can produce almost any curve,
Keplerian or not so that's not a problem for you.

George

On Wed, 12 Sep 2007 21:13:25 +0100, "George Dishman"
wrote:


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


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.

Well apart from the fact that it is not likely to be linear, I know exactly
where it goes.


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.

I don't ignore it...I just asume it is small and leave it out.
If RI is significant, we are clearly dealing with a 'medium'.

George




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, 12 Sep 2007 21:34:20 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
news
On Thu, 30 Aug 2007 17:17:24 +0100, "George Dishman"

Indeed and care must be taken for that reason
especially with local factors like absorption
by water and oxygen in the K band. These
effects are well known though, nobody ignores
them.

....but everybody seems to ignore the most critical factor..that of
variable
light speed....

People take the frequency dependence of the
refractive index into account where it has
an effect, e.g. in pulsar dispersion.

The cause doesn't matter, the shift is less than
0.01% or 0.22nm for K band when the filter is
400nm wide - completely negligible.

You cannot assume a consant emissivity for the changing surface layer
either.

The emissivity is 100% at the bottom of the
layer Henry, Kirchoff's law requires that.

Not if its temperature is continually changing.

Yes Henry, Kirchoff's law requires it.

I don't think you know what Kirschoff's Law is about.

Nor can you have photons with negative lengths...

That's your 'wave equation' of course....

No that's YOUR 'wave model', you have no wave equation.

Photons are particles.

Right, but you don't have an equation for that either.

According to my model...which works..photons are long particles carrying their
own intrinsic oscillations....oscillations that are not directly related to the
man made electric waves they often make up.


No you aren't. You didn't even consider the main factor, the temperature
gradient in the water and its affect on viscosity....
We know the ball's volume will decrease nonlinearly and we can assume it
remains in temperature equilibrium with the water.

The sea's temperature changes only slightly with
depth after the first few tens of metres, and the
effect on the ball will be minimal. Viscosity
has no effect at all on the volume of the ball.

I know that George.

Thank goodness.

I'm talking about the rate of fall, ...

No, you were talking about the ball being compressed
by pressure as an analogy for a photon being squeezed
by the differential velocity due to acceleration at
the time of emission, or some such rubbish, you never
mentioned rate of fall.

George, look back. The question was, "how does the rate of fall of a rubber
ball vary as it sinks in the ocean?".

The two important relationships are the temperature gradient in the water
(which markedly affects viscosity at near 0 C) and the variation in the bulk
modulus of rubber with pressure.

Now I don't expect anyone to actually work it out mathematically . ..but a
computer model could do it reasonably accurately.



Since the gas is a black body radiator, it must
also be a perfect absorber. As the density rises,
it becomes completely opaque which is why you
cannot see through to a second layer.

they are big assumptions...

Nope, they are results confirmed by lab tests.

George, you are very confused.

Black body radiators don't necessarily have an emissivity of 1. It can have any
value. However, according to Kischoff, the absorptivity must always equal the
emissivity at equilibrium.


program.

......so you believe that cepheid curves are Keplarian out of pure
coincidence?

No, I believe you have added so many adjustable
parameters in your program that you can fit any
curve, Keplerian or not.

George, the well known cepheid curve is Keplerian...whether you like it or
not...

Nope, but your program can produce almost any curve,
Keplerian or not so that's not a problem for you.

George, all my source velocity curves are rigidly Keplerian.

George




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, 12 Sep 2007 20:59:26 +0100, "George Dishman"
wrote:


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

...

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.

So every single photon frm a distant star is stretched to over 600 metres wide
by the time it reaches Earth?

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

....and my statistical theory explains why...

George




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, 12 Sep 2007 20:54:13 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Thu, 30 Aug 2007 10:31:32 +0100, "George Dishman"

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.

impossible. My interpretation is that the interference is caused by factors
unknown, probably different photon speeds,

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.

....the light travels a long way before it reaches the Earth's atmosphere.


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.

You have no model of a photon so how would you know?


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.

George's Giant Photons explain everything.

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

Intensity Henry.


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.

it can be caused by ADopppler, some VDoppler or shift in Planck curve.

George, quite clearly, if L Car is a huffpuff, its maximum temperture

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.

George, these exist in your dreams....

...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.

I have.


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.

George, a relativist 100LYs away would come up with all kinds of ridiculous
theories to expain our sun's apparent variation in luminosity...due solely to
c+v arising from its orbit around the sun/Jupiter barycentre. Satirn would give
it a bit of a bump too.



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.

sure


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.

far too slow...



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


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

The whole method is useless and incapable of producing anything concrete.

George




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 Wed, 12 Sep 2007 21:13:25 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...

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

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

Well apart from the fact that it is not likely to be linear, I know exactly
where it goes.

That's a laugh, Let's see the equation including it
then Henry. Be sure to include the higher order
terms if you think it is non-linear.

George

Henri Wilson wrote:
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:

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....

READ IT THEN!
You obviously haven't so far.

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.

This idiocy over and over and over!

'Coherent' isn't quite the correct expression, but since you use it
let 'coherent' in this context mean that the surface of equal phase
('wavefront') can be considered to be a plane.

Since you appear to be deaf, I will have to shout:

OF BLOODY COURSE THE LIGHT FROM DIFFERENT PARTS OF
A RESOLVED STAR ISN'T COHERENT OVER 600m!

It is the light from a _point_ (area resolution) on
the star that is coherent (plane wavefront).
The wavefronts from two different parts of the star
have an angle to each other, and is NOT coherent.

THATS WHY THEY ARE FOCUSED AT TWO DIFFERENT POINTS ON THE CCD!

Whether we are talking about a 600m telescope, or
an interferometer with two mirrors 600m apart,
WE GET AN IMAGE _BECAUSE_ THE LIGHT IS INCOHERENT!

The only kind of 'image' you can get from coherent light
is a single dot (airy disk).
we get an image _because_ the light from different
parts of the star is NOT coherent.

THE INTERFEROMETER WORKS BECAUSE THE LIGHT FROM DIFFERENT
PARTS OF THE STAR IS INCOHERENT.

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.

....and over and over and over.

Henri, you have to learn how a telescope works to
understand how an interferometer like the VLTI works.
You know neither.

Let me make a last attempt to make you understand.
Please consider the scenarios below.

If you don't bother to read it properly, don't respond.
-------------------------------------------------------

I am fed up with your autistic responses which only
reveal that you haven't read what you are responding to.

In all cases below, the observed object is l Carinae.
The angular diameter of l Carinae is ca. 3 mas.
Let us assume that lambda = 2u IR (like in the VLTI).

#1 A 8m aperture telescope:
---------------------------
The 'image' of the star is an airy disk.
http://support.svi.nl/wiki/AiryDisk
The diameter of the central disk is equivalent to 60mas.
The star is not resolved, and can be considered a point source.
That is to say - the wavefront can be considered to be a plane
over the aperture of the telescope.
(The phase difference between the wavefronts of light from
opposite sides of the star is too small to be significant.)

#2 A 600m aperture telescope:
-----------------------------
The image of the star is a disk equivalent to 3 mas.
(The diameter of the central airy disk of a point source
would be equivalent to 0.8mas. The resolution is 0.8mas)
The star is resolved.
The wavefronts from two points at the opposite sides of the star
will have an angle to each other (the light is not coherent),
and over the big aperture this angle will make a phase difference
big enough so that the two sides will be focused at different spots
on the CCD.

#3 A 600m aperture telescope consisting of 8m segments:
--------------------------------------------------------
The 'image' from each 8m segment will be an airy disk
with central disk equivalent to 60 mas.
But all the airy disks are focused on top of each other.
The interference pattern will be a disk equivalent
to 3 mas, which is identical to the image of the star in #2.

If we remove some of the segments, the image will get
'freckles', but it will still be basically the same image.

#4 A "telescope" consisting of two 8m segments 600m apart:
----------------------------------------------------------
This is an interferometer in principle like the VLTI.
It can be considered as #3 where all but two segments are removed.
The 'image' from each 8m segment will be an airy disk
with central disk equivalent to 60 mas.
These two airy disks are focused on top of each other.
The interference pattern will be a disk equivalent to 3mas.
The disk will have fringes in it, but it is still basically
the same image as in #2.

The wavefronts from two points at the opposite sides of the star
will have an angle to each other (the light is not coherent).
This means that since the segments are 600m apart, the phase
between the two wavefronts will be significantly different
in the two segments. That's why the two 60mas disks will
interfere in such a way that the streaked image of the star
appears.

Note this, Henri.
The interferometer works _because_ the light from
opposite sides of the star are sufficiently 'incoherent'
to be resolved.

But you will of course ignore this, and keep asking
me to "provide an explanation as to how light emitted
from many parts of a star's surface, can end up in
phase over a distance of 600m."

I have explained why this question reveals your
ignorance for the last time now.

Paul

On Thu, 13 Sep 2007 15:36:34 +0200, "Paul B. Andersen"
wrote:

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



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.

This idiocy over and over and over!

'Coherent' isn't quite the correct expression, but since you use it
let 'coherent' in this context mean that the surface of equal phase
('wavefront') can be considered to be a plane.

Since you appear to be deaf, I will have to shout:

OF BLOODY COURSE THE LIGHT FROM DIFFERENT PARTS OF
A RESOLVED STAR ISN'T COHERENT OVER 600m!

It is the light from a _point_ (area resolution) on
the star that is coherent (plane wavefront).
The wavefronts from two different parts of the star
have an angle to each other, and is NOT coherent.

So you are claiming that a single light quanta can have a 'cross section' of
over 600 metres by the time it reaches Earth????

THATS WHY THEY ARE FOCUSED AT TWO DIFFERENT POINTS ON THE CCD!

Whether we are talking about a 600m telescope, or
an interferometer with two mirrors 600m apart,
WE GET AN IMAGE _BECAUSE_ THE LIGHT IS INCOHERENT!

This doesn't appear to have anything to do with iinterferometry.

The only kind of 'image' you can get from coherent light
is a single dot (airy disk).
we get an image _because_ the light from different
parts of the star is NOT coherent.

THE INTERFEROMETER WORKS BECAUSE THE LIGHT FROM DIFFERENT
PARTS OF THE STAR IS INCOHERENT.

So all you are talking about is a normal image.


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.

...and over and over and over.

Henri, you have to learn how a telescope works to
understand how an interferometer like the VLTI works.
You know neither.

Let me make a last attempt to make you understand.
Please consider the scenarios below.

If you don't bother to read it properly, don't respond.
-------------------------------------------------------

I am fed up with your autistic responses which only
reveal that you haven't read what you are responding to.

In all cases below, the observed object is l Carinae.
The angular diameter of l Carinae is ca. 3 mas.
Let us assume that lambda = 2u IR (like in the VLTI).

#1 A 8m aperture telescope:
---------------------------
The 'image' of the star is an airy disk.
http://support.svi.nl/wiki/AiryDisk
The diameter of the central disk is equivalent to 60mas.
The star is not resolved, and can be considered a point source.
That is to say - the wavefront can be considered to be a plane
over the aperture of the telescope.
(The phase difference between the wavefronts of light from
opposite sides of the star is too small to be significant.)

#2 A 600m aperture telescope:
-----------------------------
The image of the star is a disk equivalent to 3 mas.
(The diameter of the central airy disk of a point source
would be equivalent to 0.8mas. The resolution is 0.8mas)
The star is resolved.
The wavefronts from two points at the opposite sides of the star
will have an angle to each other (the light is not coherent),
and over the big aperture this angle will make a phase difference
big enough so that the two sides will be focused at different spots
on the CCD.

Hahahahaha! You must be kidding, surely....
How can incoherent light from two entirely different sources create
interference? It's probably not even the same wavelength..and who know when it
was emitted....

Hahahoohohahahah!


#3 A 600m aperture telescope consisting of 8m segments:
--------------------------------------------------------
The 'image' from each 8m segment will be an airy disk
with central disk equivalent to 60 mas.
But all the airy disks are focused on top of each other.
The interference pattern will be a disk equivalent
to 3 mas, which is identical to the image of the star in #2.

If we remove some of the segments, the image will get
'freckles', but it will still be basically the same image.

#4 A "telescope" consisting of two 8m segments 600m apart:
----------------------------------------------------------
This is an interferometer in principle like the VLTI.
It can be considered as #3 where all but two segments are removed.
The 'image' from each 8m segment will be an airy disk
with central disk equivalent to 60 mas.
These two airy disks are focused on top of each other.
The interference pattern will be a disk equivalent to 3mas.
The disk will have fringes in it, but it is still basically
the same image as in #2.

The wavefronts from two points at the opposite sides of the star
will have an angle to each other (the light is not coherent).
This means that since the segments are 600m apart, the phase
between the two wavefronts will be significantly different
in the two segments. That's why the two 60mas disks will
interfere in such a way that the streaked image of the star
appears.

Note this, Henri.
The interferometer works _because_ the light from
opposite sides of the star are sufficiently 'incoherent'
to be resolved.

No Paul, an interference pattern of some kind just happens to appear because
all the light from the star is unified in speed as it travels across space
(and maybe phase aligned as well).
The interference varies with c+v as the star orbits .

But you will of course ignore this, and keep asking
me to "provide an explanation as to how light emitted
from many parts of a star's surface, can end up in
phase over a distance of 600m."

I have explained why this question reveals your
ignorance for the last time now.

You really should study some genuine physics Paul....

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, 13 Sep 2007 15:36:34 +0200, "Paul B. Andersen"
wrote:

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


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.
This idiocy over and over and over!

'Coherent' isn't quite the correct expression, but since you use it
let 'coherent' in this context mean that the surface of equal phase
('wavefront') can be considered to be a plane.

Since you appear to be deaf, I will have to shout:

OF BLOODY COURSE THE LIGHT FROM DIFFERENT PARTS OF
A RESOLVED STAR ISN'T COHERENT OVER 600m!

It is the light from a _point_ (area resolution) on
the star that is coherent (plane wavefront).
The wavefronts from two different parts of the star
have an angle to each other, and is NOT coherent.

So you are claiming that a single light quanta can have a 'cross section' of
over 600 metres by the time it reaches Earth????

I am claiming that the diameter of the aperture strongly
affects the probability for where the photon will hit
at the CCD. The photon from a point source will with high
confidence hit within a diameter proportional to
the wavelength divided by the aperture diameter.

If you think that implies that the photon must have
a cross section equal to the aperture, then it illustrates
how hopeless idiotic your idea of a photon is.

This response is typical for you.
Effectually you claim that big telescopes do not have
better resolution than small ones, because you cannot
understand how the size of the apperture can affect a photon.

Since you drop off at the very beginning,
there is no point in going on.
This is obviously way beyond your abilities.

[snip]

Paul

"Henri Wilson" HW@.... wrote in message
...
On Wed, 12 Sep 2007 20:59:26 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
...

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.

So every single photon frm a distant star is stretched to over 600 metres
wide
by the time it reaches Earth?

That is for you to decide, you have to explain how
an individual photon's behaviour can depend on the
paths through both telescopes, Paul and I are only
telling you what the observed result is. It is just
a larger example of Young's Slits so you shouldn't
be too surprised.

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

...and my statistical theory explains why...

More lies Henry? You don't have a statistical theory,
only Ritz's macroscopic velocity equation and my
extension for speed equalisation. In fact, come to
think of it, you have never originated anything for
yourself.

George