Why are the 'Fixed Stars' so FIXED?

On Wed, 19 Sep 2007 21:17:09 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 17 Sep 2007 19:33:27 +0100, "George Dishman"

wrote:


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


Your model consists of two equations:

v = u+c

and

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

Neither has any oscillatory term.


Now write the equation for the wave in a moving organ pipe..whilst it is
producing sound...

Why would I do that? It isn't a solution to either
of the equations so not part of your theory.

You obviously don't understand my photon theory..

You don't have a photon theory, you only talk
about them as bursts of classical waves but
your equations do not produce photons. See
my description of your 'EM' program for an
analysis and then think what you could do to
your equations to make such a pattern self
sustaining during propagation. I don't think
it is possible.

See my reply to Ghost in the other thread.

....oscillations that are not directly related to the
man made electric waves they often make up.

Then you need some equation that links the two,
your model is again lacking.

I don't need an equation. I can imagine the result.

Then you have no theory, you just imagine you
have one.

I admit it isn't complete..

See my response to your new post. If you have any
new equations to add, let me know but the summary
is based on the two you have.

The definition of a "black body" is that it
absorbs all incident radiation. That means
absorptivity = 1 and by Kirschoff's Law that
means emissivity = 1.

Yes well Ok ,

OK, so think twice before saying I am confused next
time, you have been wrong every time.

that's not what I meant to convey. What I meant was "emitters can
have a black body type curve but an emissivity of less than 1".

"Gray atmosphere" I believe is the term in this
field, but Kirchoff's Law still applies and since
the opacity increases with depth you still cannot
see through to deeper layers beyond the photosphere.

But you just asked how long it took light from the sun's core to reach the
surface. If t reaches the surface at all, surely you can 'see' it.

You see the surface from which it was last emitted.
Think of an opaque film with light shining on it.
The film heats up and emits black-body radiation
so a black plastic film with sunlight might glow
at 300K even though the spectrum illuminating it
is over 5000K. You can still feel the warmth but
there will be a slight delay if a shaow passes
over it. Try it with a black plastic bag in a frame
in your back garden.

Not convincing....

While on that subject. I once decided to heat my computer room in winter by
hanging a black plastic curtin over the large north facing window ( I live in
OZ). This worked well, acting as both a re-radiator and a double
insulator...The room warmed very efficiently and was dark enough for me to see
the screen.
There was only one problem...after several weeks, I started to get a sore
throat whenever I sat there for long. I finally woke up to the fact that the
plastic was slowly decaying and giving off what was probably a pretty deadly
gas...It didn't kill me but I wouldn't recommend it to anyone else.

I later rigged up a similar curtain using sheet metal and achieved the same
result....plenty of warmth with little bright sunlight...



Perhaps, I have yet to see any verification test
results for that, but you allow for harmonics and
contributions from a second source which mean the
total is not Keplerian.

My program also incorporates 'egg shaped stars', for instance in tidal
lock.
This is another possible source of an apparent 'overtone'

Right, nothing to do with a Keplerian orbit at all.
Now you understand the point.

...the curve shape is still Keplerian....

Nope, you have extras, it is just arbitrary but
no matter, we know it isn't an orbit anyway.

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, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 17 Sep 2007 19:23:54 +0100, "George Dishman"

Multiple images starts at what I called the critical
distance given by d = c^2/a where a is the peak radial
acceleration. Variation of 1.5 mags means a ratio of
4:1 in luminosity which means the speed unification
distance is about 75% of the critical distance. Other
values are

Percentage of
Mag 1:n critical
1 2.5 60.19%
1.5 4.0 74.88%
2 6.3 84.15%
3 15.8 93.69%
4 39.8 97.49%
5 100.0 99.00%
6 251.2 99.60%
7 631.0 99.84%
8 1584.9 99.94%
9 3981.1 99.97%

Isn't it magical how we see variations of up to 9 mags
in some stars yet NEVER see multiple images.

It is true some stars are reported to vary by 7-9 mags. This cannot be
explained solely by c+v effects.

Of course it can Henry, that's the point. Suppose
we take an arbitrary figure of 10 light years
for the speed equalisation distance in the space
surrounding some star. To get 1.5 mag variation
you need a critical distance of 13.35 light years.
To get 9 mag that needs to be 10.0025 light years,
just a 25% increase in the peak orbital
acceleration.

What the hell are you talking about?

Magnitude changes of 2-3 are easy to produce without the appearance of
peaks in
the brightnes curves.

Sure, but there is no law that says stars can't
orbit a little faster or have slightly higher
eccentricity so your problem is explaining why
we _don't_ see such spikes. The answer of course
is that every sysytem we see only shows VDoppler,
never ADoppler.

You get 10/10 for stubborness.

Even more
remarkable when you realise the critical distance depends
on inclination while the speed equalisation depends only
on the nature of the ISM, is that the heliosphere?

I have explained the role of inclination.

You miss the point. Suppose a star shows 5 mags
variation to astronomers on Earth and the orbit
has an inclination of 45 degrees. An astronomer
on a nearby star might see an inclination of
44.44 degrees, and he would see a variation of
9 mags because of the slightly higher peak
acceleration. At 44.42 degrees, he sees multiple
images. There is no reason why that shouldn't be
us, you need fairies to carefullt orient all the
inclications in the galaxy so that they have just
the right angle to little old Earth :-)


....and this by a person who boasts about his maths ability....
How embarassing for you....

You can use edge on orbits as long as
you remember that the velocities and accelerations used are really v and a
multiplied by cos(pitch) .

I believe unification is a two stage affair....one close to the star and
the
other much smaller in rate and over large distances of space.

You don't know what you are saying half the time....

You don't understand basic arithmetic!

The computer does the arithmetic.

Sure, all my numbers above were done in a few
minutes with a spreadsheet, but you couldn't
write that and don't understand the consequences
because you don't practice your maths.

Your numbers are completely wrong...
I think you have confused logs with exponentials...

I just look at the curves and figures.

Sure henry, pretty pictures and colouring books
are OK, but you need to do the maths if you want
to play with real physics.

Its much quicker than fiddling about with c^2/a.

Sure Henry, sure.

What a remarkable coincidence, or is it the fairies
again?

No.. all the fairies are on Einstein's side...

Nope, all Einstein needs is Pythagoras. You should
learn it some time.

Einstein used Pythagoras to deduce that raindrops take longer to reach
the
ground when viewed through the window of his moving train.

It's all beyond your level of maths Henry, don't
worry about it.

I wont George.
So do the randrops take longer to reach the ground?

Don't worry about it Henry, it's too complex for
you. See if you can follow what I said about
inclination instead.

...not one of your better posts George....You probably need another holiday...

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, 19 Sep 2007 21:45:03 +0100, "George Dishman"
wrote:


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


Garbage, you get one photon from a distant nebula
or whatever and it lands on the CCD at a point that
depends on the whole telescope. A second later you
get the next photon and it lands on a different
pixel because it came from a different part of the
nebula a few light years from the first photon.

Well you wont get a continuous wavefront or any interference then...You
just
get a normal image...albeit very slowly..

That's what you might think but if you have
the two separated telescopes, you still get
interference. Henry, I'm just telling you
what happens in practice, it is up to you
to make sure your ideas can cope with that.

George, I don't think you have a clue.

What do you think might be the focal length of a 100 metre aperture
telescop?..maybe 1000 metres??

Do you believe a some kind of mirror system is erected way up in the sky and
moved around with a tracking mechanism?

Of course not.

The fact is the two separated telescopes don't move like a single mirror would.
Each is rotated on its own base and has its own 'eyepiece'. The two behave
somewhat like adjacent lines on a grating. The incoming signals are then
analysed in terms of the phase difference between the two images.

What "common wavefront" are you talking about?

...that across a spherical surface normal to the LOS.

OK, that's what you get with each individual
photon.

So photons spread out as they travel... I can go along with that to some
extent....but there is still a problem with gammas...

Otherwise, no deteiled image could be formed at all.

Each photon lands on a pixel that depends on the
source location and the PSF, nothing else. If it
was affected by the previous photon in some way
then photons coming from one part of the nebula
would be scattered over the CCD depending on where
the preceding photn was from, the image would be
blurred. You only get a sharp image because every
photon behaves completely independently of all
other photons and goes where it is supposed to.

Well I would be inclined to agree with that...and c+v doesn't cause any
blurring because no photons overtake other ones.

It wouldn't matter if they did, as long as they don't
alter the fact that the wavefront is normal to the
line of sight, you get an image.

.....an image of events that occurred at different times.

However this doesn't back up your and Paul's claim that one photon can
extend
and remain coherent over a 600 m circle.

No, that is quite different. The evidence is that
combining the light from widely separated telescopes
produces interference patterns. What the imaging
argumnent requires is that you cannot combine multiple
photons into a single 'average' wavefront because each
must point independently back to its source or the
image would collapse to a point. Low arrival rates
also mean you can directly see interference in systems
where the detectors work in photon counting mode, just
as with PM tubes and gratings. All the evidence says
the same thing.

You're guessing George....see above....

[ gamma rays ...]
It would be quite dangerous...and we're being bombarded by them
continuously...

It would be if we were, but the ozone layer stops them.

Not all...

Bye bye then.

well there is probably more truth in that than you realise.

That's why people are worried, but you have your
argument the wrong way round, gamma rays are
dangerous because they dump all their energy
into a tiny space, typically one atom, regardless
of how wide the wavefront is. That's why we know
they are particles, not classical waves.

....and if you move towards a visible light source at 0.99c, you'll probably run
into quite a few gammas..

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" HW@.... wrote in message
...
On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
On Mon, 17 Sep 2007 19:23:54 +0100, "George Dishman"

Multiple images starts at what I called the critical
distance given by d = c^2/a where a is the peak radial
acceleration. Variation of 1.5 mags means a ratio of
4:1 in luminosity which means the speed unification
distance is about 75% of the critical distance. Other
values are

Percentage of
Mag 1:n critical
1 2.5 60.19%
1.5 4.0 74.88%
2 6.3 84.15%
3 15.8 93.69%
4 39.8 97.49%
5 100.0 99.00%
6 251.2 99.60%
7 631.0 99.84%
8 1584.9 99.94%
9 3981.1 99.97%

Isn't it magical how we see variations of up to 9 mags
in some stars yet NEVER see multiple images.

It is true some stars are reported to vary by 7-9 mags. This cannot be
explained solely by c+v effects.

Of course it can Henry, that's the point. Suppose
we take an arbitrary figure of 10 light years
for the speed equalisation distance in the space
surrounding some star. To get 1.5 mag variation
you need a critical distance of 13.35 light years.
To get 9 mag that needs to be 10.0025 light years,
just a 25% increase in the peak orbital
acceleration.

What the hell are you talking about?

Check the numbers, see for yourself. For a given
value of peak acceleration, draw a graph of peak
variation as a function of distance.

Magnitude changes of 2-3 are easy to produce without the appearance of
peaks in
the brightnes curves.

Sure, but there is no law that says stars can't
orbit a little faster or have slightly higher
eccentricity so your problem is explaining why
we _don't_ see such spikes. The answer of course
is that every sysytem we see only shows VDoppler,
never ADoppler.

You get 10/10 for stubborness.

Do the arithemetic and see for yourself.

Even more
remarkable when you realise the critical distance depends
on inclination while the speed equalisation depends only
on the nature of the ISM, is that the heliosphere?

I have explained the role of inclination.

You miss the point. Suppose a star shows 5 mags
variation to astronomers on Earth and the orbit
has an inclination of 45 degrees. An astronomer
on a nearby star might see an inclination of
44.44 degrees, and he would see a variation of
9 mags because of the slightly higher peak
acceleration. At 44.42 degrees, he sees multiple
images. There is no reason why that shouldn't be
us, you need fairies to carefullt orient all the
inclications in the galaxy so that they have just
the right angle to little old Earth :-)

...and this by a person who boasts about his maths ability....
How embarassing for you....

Check the numbers then, they are correct. You
have never grasped this problem because you
don't do the arithmetic.

.. all my numbers above were done in a few
minutes with a spreadsheet, but you couldn't
write that and don't understand the consequences
because you don't practice your maths.

Your numbers are completely wrong...
I think you have confused logs with exponentials...

I think you are forgetting that magnitudes
are logarithmic, or perhaps that ADoppler
goes as 1/(c^2-da). As d approaches c^2/a,
you get a very rapid rise because the
denominator approaches zero.

..not one of your better posts George....You probably need another
holiday...

It is all correct.

George

In article ,
George Dishman wrote:
"Henri Wilson" HW@.... wrote in message
...

.......................

Sure, but there is no law that says stars can't
orbit a little faster or have slightly higher
eccentricity so your problem is explaining why
we _don't_ see such spikes. The answer of course
is that every sysytem we see only shows VDoppler,
never ADoppler.

You get 10/10 for stubborness.

Do the arithemetic and see for yourself.

Now you get 20/10 for stubbornness.... :-)

And you're absolutely right, of course!

--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://stjarnhimlen.se/

On 21 Sep, 00:36, HW@....(Henri Wilson) wrote:
On Wed, 19 Sep 2007 21:45:03 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...

Garbage, you get one photon from a distant nebula
or whatever and it lands on the CCD at a point that
depends on the whole telescope. A second later you
get the next photon and it lands on a different
pixel because it came from a different part of the
nebula a few light years from the first photon.

Well you wont get a continuous wavefront or any interference then...You
just get a normal image...albeit very slowly..

That's what you might think but if you have
the two separated telescopes, you still get
interference. Henry, I'm just telling you
what happens in practice, it is up to you
to make sure your ideas can cope with that.

George, I don't think you have a clue.

Let's see ...

What do you think might be the focal length of a 100 metre aperture
telescop?..maybe 1000 metres??

MELIPAL and ANTU are 8.2m in diameter and have a focal
length of 14.35m.

Do you believe a some kind of mirror system is erected way up in the sky and
moved around with a tracking mechanism?

Of course not.

http://tinyurl.com/3dybf3

The position of the secondary mirror above the primary
and location of the Nasmyth and Coude focal planes are
shown in the diagram, why didn't you look instead of
making a fool of yourself.

The fact is the two separated telescopes don't move like a single mirror would.
Each is rotated on its own base and has its own 'eyepiece'. The two behave
somewhat like adjacent lines on a grating.

Exactly, which is why I told you that your ballistic
grating equation applies.

The incoming signals are then
analysed in terms of the phase difference between the two images.

Yes, the two paths are brought together so that the phase
difference creates an interference pattern as shown in
the ESO diagram.

What "common wavefront" are you talking about?

...that across a spherical surface normal to the LOS.

OK, that's what you get with each individual
photon.

So photons spread out as they travel... I can go along with that to some
extent....but there is still a problem with gammas...

Individual photons exhibit interference over the 102m
between the two telescopes. Gamma telescopes are smaller
but if they could be made to a comparable tolerance
(fraction of a wavelength) they would show the same
behaviour. If that gives a problem, it is your problem.

Otherwise, no deteiled image could be formed at all.

Each photon lands on a pixel that depends on the
source location and the PSF, nothing else. If it
was affected by the previous photon in some way
then photons coming from one part of the nebula
would be scattered over the CCD depending on where
the preceding photn was from, the image would be
blurred. You only get a sharp image because every
photon behaves completely independently of all
other photons and goes where it is supposed to.

Well I would be inclined to agree with that...and c+v doesn't cause any
blurring because no photons overtake other ones.

It wouldn't matter if they did, as long as they don't
alter the fact that the wavefront is normal to the
line of sight, you get an image.

....an image of events that occurred at different times.

"Each photon" Henry. Telescopes detect individual
photons.

However this doesn't back up your and Paul's claim that one photon can
extend
and remain coherent over a 600 m circle.

No, that is quite different. The evidence is that
combining the light from widely separated telescopes
produces interference patterns. What the imaging
argumnent requires is that you cannot combine multiple
photons into a single 'average' wavefront because each
must point independently back to its source or the
image would collapse to a point. Low arrival rates
also mean you can directly see interference in systems
where the detectors work in photon counting mode, just
as with PM tubes and gratings. All the evidence says
the same thing.

You're guessing George....see above....

It is obvious that, if you average the location of
the wavefronts of two photons, the normal points
back to a location half way between the two sources,
you could not form a sharp image with a telescope
which Paul has been pointing out for some time.

Photon counting mode is used regularly in astronomy
and long exposures to handle low photon arrival
rates are commonplace. I am telling you what happens,
writing the equations to explain it is your problem.

George

"Henri Wilson" HW@.... wrote in message
...
On Wed, 19 Sep 2007 21:17:09 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
On Mon, 17 Sep 2007 19:33:27 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
m...
[I wrote:]

Your model consists of two equations:

v = u+c

and

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

Neither has any oscillatory term.


Now write the equation for the wave in a moving organ pipe..whilst it
is
producing sound...

Why would I do that? It isn't a solution to either
of the equations so not part of your theory.

You obviously don't understand my photon theory..

You don't have a photon theory, you only talk
about them as bursts of classical waves but
your equations do not produce photons. See
my description of your 'EM' program for an
analysis and then think what you could do to
your equations to make such a pattern self
sustaining during propagation. I don't think
it is possible.

See my reply to Ghost in the other thread.

I haven't seen a new message from you to him since
this was posted so you'll need to give me a link
or summarise.

"Gray atmosphere" I believe is the term in this
field, but Kirchoff's Law still applies and since
the opacity increases with depth you still cannot
see through to deeper layers beyond the photosphere.

But you just asked how long it took light from the sun's core to reach
the
surface. If t reaches the surface at all, surely you can 'see' it.

You see the surface from which it was last emitted.
Think of an opaque film with light shining on it.
The film heats up and emits black-body radiation
so a black plastic film with sunlight might glow
at 300K even though the spectrum illuminating it
is over 5000K. You can still feel the warmth but
there will be a slight delay if a shaow passes
over it. Try it with a black plastic bag in a frame
in your back garden.

Not convincing....

Not meant to be, simply an analogy to aid
understanding. For proof, you need to calculate
the mean free path.

While on that subject. I once decided to heat my computer room in winter
by
hanging a black plastic curtin over the large north facing window ( I live
in
OZ). This worked well, acting as both a re-radiator and a double
insulator...The room warmed very efficiently and was dark enough for me to
see
the screen.
There was only one problem...after several weeks, I started to get a sore
throat whenever I sat there for long. I finally woke up to the fact that
the
plastic was slowly decaying and giving off what was probably a pretty
deadly
gas...It didn't kill me but I wouldn't recommend it to anyone else.

Nasty.

I later rigged up a similar curtain using sheet metal and achieved the
same
result....plenty of warmth with little bright sunlight...

Nice. Now imagine a material which is slighly
translucent but so thick no light gets through,
just the heat. You might be able to see 2mm into
it but it is 100cm thick. That's how the surface
of the Sun behaves, except that in addition it
is more transparent near the surface and gets
denser with depth, say 1cm of transparent before
you get to the 1mm translucent layer. The far side
is hot where the Sun hits it, the near side is
cooler because of the insulating effect of the
material, and looking into the surface you see a
small temperature gradient from about 11 mm into
the material.

George

On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
...
On Mon, 17 Sep 2007 19:23:54 +0100, "George Dishman"

Multiple images starts at what I called the critical
distance given by d = c^2/a where a is the peak radial
acceleration. Variation of 1.5 mags means a ratio of
4:1 in luminosity which means the speed unification
distance is about 75% of the critical distance. Other
values are

Percentage of
Mag 1:n critical
1 2.5 60.19%
1.5 4.0 74.88%
2 6.3 84.15%
3 15.8 93.69%
4 39.8 97.49%
5 100.0 99.00%
6 251.2 99.60%
7 631.0 99.84%
8 1584.9 99.94%
9 3981.1 99.97%

Isn't it magical how we see variations of up to 9 mags
in some stars yet NEVER see multiple images.

It is true some stars are reported to vary by 7-9 mags. This cannot be
explained solely by c+v effects.

Of course it can Henry, that's the point. Suppose
we take an arbitrary figure of 10 light years
for the speed equalisation distance in the space
surrounding some star. To get 1.5 mag variation
you need a critical distance of 13.35 light years.
To get 9 mag that needs to be 10.0025 light years,
just a 25% increase in the peak orbital
acceleration.

What the hell are you talking about?

Check the numbers, see for yourself. For a given
value of peak acceleration, draw a graph of peak
variation as a function of distance.

What do you think my program does?

Magnitude changes of 2-3 are easy to produce without the appearance of
peaks in
the brightnes curves.

Sure, but there is no law that says stars can't
orbit a little faster or have slightly higher
eccentricity so your problem is explaining why
we _don't_ see such spikes. The answer of course
is that every sysytem we see only shows VDoppler,
never ADoppler.

You get 10/10 for stubborness.

Do the arithemetic and see for yourself.

the computer does it...

Even more
remarkable when you realise the critical distance depends
on inclination while the speed equalisation depends only
on the nature of the ISM, is that the heliosphere?

I have explained the role of inclination.

You miss the point. Suppose a star shows 5 mags
variation to astronomers on Earth and the orbit
has an inclination of 45 degrees. An astronomer
on a nearby star might see an inclination of
44.44 degrees, and he would see a variation of
9 mags because of the slightly higher peak
acceleration. At 44.42 degrees, he sees multiple
images. There is no reason why that shouldn't be
us, you need fairies to carefullt orient all the
inclications in the galaxy so that they have just
the right angle to little old Earth :-)

...and this by a person who boasts about his maths ability....
How embarassing for you....

Check the numbers then, they are correct. You
have never grasped this problem because you
don't do the arithmetic.

.....and I really thought you understood this stuff....
Unification ensures we wont see multiple images....we wont even see magnitude
changes due to BaTh above about .

.. all my numbers above were done in a few
minutes with a spreadsheet, but you couldn't
write that and don't understand the consequences
because you don't practice your maths.

Your numbers are completely wrong...
I think you have confused logs with exponentials...

I think you are forgetting that magnitudes
are logarithmic, or perhaps that ADoppler
goes as 1/(c^2-da). As d approaches c^2/a,
you get a very rapid rise because the
denominator approaches zero.

the situation never arises...


George


Henri Wilson. ASTC,BSc,DSc(T)

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

"Dr. Henri Wilson" HW@.... wrote in message
...
On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
m...
On Mon, 17 Sep 2007 19:23:54 +0100, "George Dishman" wrote:

Multiple images start at what I called the critical
distance given by d = c^2/a where a is the peak radial
acceleration. Variation of 1.5 mags means a ratio of
4:1 in luminosity which means the speed unification
distance is about 75% of the critical distance. Other
values are

Percentage of
Mag 1:n critical
1 2.5 60.19%
1.5 4.0 74.88%
2 6.3 84.15%
3 15.8 93.69%
4 39.8 97.49%
5 100.0 99.00%
6 251.2 99.60%
7 631.0 99.84%
8 1584.9 99.94%
9 3981.1 99.97%

Isn't it magical how we see variations of up to 9 mags
in some stars yet NEVER see multiple images.

It is true some stars are reported to vary by 7-9 mags. This cannot be
explained solely by c+v effects.

Of course it can Henry, that's the point. Suppose
we take an arbitrary figure of 10 light years
for the speed equalisation distance in the space
surrounding some star. To get 1.5 mag variation
you need a critical distance of 13.35 light years.
To get 9 mag that needs to be 10.0025 light years,
just a 25% increase in the peak orbital
acceleration.

What the hell are you talking about?

Check the numbers, see for yourself. For a given
value of peak acceleration, draw a graph of peak
variation as a function of distance.

What do you think my program does?

If you think it can tell you the above results,
use it. Tell me what distance as a percentage
of (c^2/a) you get a variation peak variation
of 8 magnitudes (peak-to-peak will be about
0.75 magnitudes more).

Tell me whether you would get multiple images if
the distance were 1% greater.

If that is at an inclination of 45 degrees, would
we see multiple images if it were 1 degree closer
to edge-on?

Your numbers are completely wrong...

Then use your program and post your alternatives.

George

On Wed, 19 Sep 2007 22:16:11 +0100, "George Dishman"
wrote:


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


Incidentally, ANTU and MELIPAL are only 102m apart
if you check the diagram, but the point remains, the
inteferometer is still capable of detecting individual
photons and intereference results from the _two_
telescopes.

You are not making it clear whethr the result is a normal image or an
interfrence pattern. All I would expect to get from this greater
resolution is
a normal image.

The system was used to measure L Car. Using just one of
the two telescopes, you get an image which is basically
a point, the size of the imaged star will be less than
the PSF (point spread function) of the scope which is
set by the diameter of the dish. If you could make a
reflector 100m across, that diffraction limit would be
less than the image diameter so you could resolve the
star as a disc. the system goes part way there by
combining the light from the two telescopes. If the
star were static, the interference would modify the
detector output like putting a photocell at a fixed
point in a Young's Slits experiment, say 50%. However,
the earth rotates and as it does so it sweeps across
the star so essentially the interference pattern is
swept over the detector as the star's image appears
to moves across the sky. That allows the contrast
ratio between the dark and light fringes to be seen
as a variation in the signal from the detector and
the contrast ratio depends on the star's diameter.

If as you claim, individual photons DO spread themselves out...or
disperse.. as
they travel, then this fact would give a litttle more information to work
on in
constructing a photon model. It doesn't really conflict with my traveling
wave
idea. It simply means the whole crossection becomes larger.

You have to think of your diagram as a sort of cross
section. The spherical wavefronts you mentioned would
look like alternate red and blue umberellas if you did
a 3D version of your program.

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

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

That is not an acceptible theory.


explain how you can get interference effects with
single photon sources like dim stars. You don't
seem to appreciate that many advanced telescopes
work in this low photon rate mode these days.

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

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

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

Regardless of your lack of understanding of that
point, even with classical waves, it should be
obvious that only the speed from x to y affects
the relative phase at the detector provided the
speed from the mirrors to the detector is the same
in both cases.
George, you are referring to a photon that acts over a distance of at
least 120
metres and somehow creates fringes.
Where does 'wavelength' come into this?

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

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

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

You know what? I don't think you know what you are talking about George.

I know what _I_ am talking about, it's the words
you try to put in my mouth that are nonsense.

You don't even know what a photon is so why are you making such stupid
claims.
Until now, you have talked nonstop about wave theory...

We have been talking about YOUR theory which is
a purely classical wave model.

now suddenly you want to
ACCEPT THAT PHOTONS ARE DISCRETE PARTICLES.
....make up your mind george.

You know perfectly well that the conventional
model is QED which is a particle based quantum
theory, I have told you that repeatedly. Remember
our recent chat about the photoelectric effect
regarding Sean's views?

QED says photons are single particles and all the
evidence bears that out. I thought you agreed with
that interpretation. It is a well known fact that
interferometers and telescopes work perfectly well
when the source is so dim that only a few photons
a second can be collected.

They can create a normal image....why shouldn't they....but where does
interference enter the equation?

An image forms because the point to which the light
reflects from a mirror is the point where the
wavefront re-radiated from all points on the
surface of the mirror produces constructive
interference. For a moment, imagine that every
atom on the surface received the wavefront and
immediately re-radiated it isotropically. Work
out what you get at some point away from the
point of the image. That is the basis of Huygens
method.

You have to decide how
your theory will handle that. QED says the
probability of the photon being detected at any
point is related to the integral of the relative
phase over all possible paths from the source to
that point.

Probability doesn't work with a sample size of unity.

You are integrating over a surface and for
all possible locations where the photon might
land. The total integral is 1, it has to go
somewhere, but for a telescope with a CCD,
there are thousands of pixels it might hit.
For a sharp image the probability is a few
adds up to nearly 1 while all the rest have a
low probability, getting smaller as you move
away from the centre of the image. That
distribution is the PSF of the system.

Your theory consists of only two equations, one
of them written by me. Neither equation explains
the existence of interference in any form
whatsoever.

George, you have NO theory.

Sorry Henry, I have QED.

It is a statistical thery that attempts to produce macroscopic results. We
are
talking about the micro PHYSICAL mode that applies to individual photon
PARTICLES..

You have no idea what sort of theory it is,
it works just fine for individual photons.

Of course you will. Individual photons have a 'cross section' that
extends
to
infinity....

Wow, Henry I'm impressed. OK my diagram shows a
side view of that cross-section.

but its 'strength' drops off rather quickly with distance.

That will depend on the source.

...or rather, the energy....

For example, a hand torch has a low probability
of sending photons back to the user if he points
it away from himself. It really is quite obvious.

I doubt
if single photon that has traveled 1700 LYs will be detectable on both
sides of
a 600 metre circle.

The longest baseline used by the GSI setup is 12400.5 km
so photons have to be bigger than that :-)

It is possible...but you wont produce any interference involving photons
emitted from both sides of a star in this way.

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

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



Do you think I have all day...

I don't think your have that much life left, but
that's what proper models do, you can find the
result all over the web. "How long does it take
for light from the Sun's core to reach us?"

they say it takes thousands of years....

Right, that's from the core. In Cepheids, most
of that time is reaching the unstable region
but it still takes some more time to reach the
photosphere hence there is a lag.

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

Your 'spheres' nonsense says the light leaves the
sphere around the star at the same speed for both
sides, but even that is irrelevant anyway.

Nice to know you understand my spheres theory.

Sure, everyone else calls it the heliosphere

http://en.wikipedia.org/wiki/Image:Solarmap.gif

http://en.wikipedia.org/wiki/Heliopause#Heliopause

OK it all adds weight to my theory...It appears I discovered the
heliopause
quite independently and without sending up any rockets.
I hope you are impressed George.

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

Sagnac disproves SR. Paul Andersen clearly showed that.

George


Henri Wilson. ASTC,BSc,DSc(T)

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