Why are the 'Fixed Stars' so FIXED?

On Sun, 23 Sep 2007 12:33:14 +0100, "George Dishman"
wrote:


"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


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?

It never gets to that point.
I have suggested before that the unification rate in any defineable volume of
space is dependent on the difference between the actual light speed and the
natural equilibrium speed in that region.

Your numbers are completely wrong...

Then use your program and post your alternatives.

You can use it. It's quite user friendly now.

www.users.bigpond.com/hewn/variables.exe

George


Henri Wilson. ASTC,BSc,DSc(T)

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

On Fri, 21 Sep 2007 15:46:12 +0100, "George Dishman"
wrote:


"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
om...
[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.

Yes of course... but outward energy flow has to be consistent...

I once had a job taking very fine photos of the sun, mainly in the H red.

The edges of the ball were quite sharp.
However, that doesn't mean a significant thick tansparent layer does not
exist....and that layer emits plenty of light.

George


Henri Wilson. ASTC,BSc,DSc(T)

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

On 23 Sep, 22:09, HW@....(Henri Wilson) wrote:
On Sun, 23 Sep 2007 12:33:14 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:kigbf317ahrqpdeknk5lrk0rsv0t7oc9f3@4a x.com...
On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message

restoring the subject from which Henry is trying
to run away

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?

It never gets to that point.

Stop ducking the issue Henry, you claimed my numbers
were wrong so prove it, put up or shut up.

Your numbers are completely wrong...

Then use your program and post your alternatives.

You can use it. ...

Why, don't you know how? I think you are so hopeless
that even this trivial arithmetic is beyond you, but
you claimed my numbers were wrong so it is up to you
to back that up, I'm not going to do it for you. Oh
and if your program gives something other than those
values, you will still need to prove it is your code
that is right and not my numbers, IMO your code is
highly suspect. Bear in mind the numbers are peak,
not peak-to-peak which I think is what your program
usually produces as the summary.

George

On Mon, 24 Sep 2007 05:53:18 -0700, George Dishman
wrote:

On 23 Sep, 22:09, HW@....(Henri Wilson) wrote:
On Sun, 23 Sep 2007 12:33:14 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:kigbf317ahrqpdeknk5lrk0rsv0t7oc9f3@4a x.com...
On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message

restoring the subject from which Henry is trying
to run away

On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
om...
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?

It never gets to that point.

Stop ducking the issue Henry, you claimed my numbers
were wrong so prove it, put up or shut up.

George, use a computer program ....and stop wasting your time with equations.

Your numbers are completely wrong...

Then use your program and post your alternatives.

You can use it. ...

Why, don't you know how? I think you are so hopeless
that even this trivial arithmetic is beyond you, but
you claimed my numbers were wrong so it is up to you
to back that up, I'm not going to do it for you. Oh
and if your program gives something other than those
values, you will still need to prove it is your code
that is right and not my numbers, IMO your code is
highly suspect. Bear in mind the numbers are peak,
not peak-to-peak which I think is what your program
usually produces as the summary.

George, I'm way ahead of you on this...
You don't even understand the basics...

George

Henri Wilson. ASTC,BSc,DSc(T)

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

On 24 Sep, 23:19, HW@....(Dr. Henri Wilson) wrote:
On Mon, 24 Sep 2007 05:53:18 -0700, George Dishman wrote:
On 23 Sep, 22:09, HW@....(Henri Wilson) wrote:
On Sun, 23 Sep 2007 12:33:14 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:kigbf317ahrqpdeknk5lrk0rsv0t7oc9f3@4a x.com...
On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message

restoring the subject from which Henry is trying
to run away

On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
om...
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?

It never gets to that point.

Stop ducking the issue Henry, you claimed my numbers
were wrong so prove it, put up or shut up.

George, use a computer program ....

Stop ducking Henry, you said my numbers were wrong
but the truth is you haven't a clue how to produce
them yourself.

and stop wasting your time with equations.

The equations told me the answers in seconds
while you are incapable of working them out
at all. You are incompetent at even trivial
arithmetic, and the only use for the computer
is to have it calculate the equation for you
anyway (which is what I did with Excel of
course).

Your numbers are completely wrong...

Then use your program and post your alternatives.

You can use it. ...

Why, don't you know how? I think you are so hopeless
that even this trivial arithmetic is beyond you, but
you claimed my numbers were wrong so it is up to you
to back that up, I'm not going to do it for you. Oh
and if your program gives something other than those
values, you will still need to prove it is your code
that is right and not my numbers, IMO your code is
highly suspect. Bear in mind the numbers are peak,
not peak-to-peak which I think is what your program
usually produces as the summary.

George, I'm way ahead of you on this...

No Henry, you can't even do a simple sum, turning
astronomical magnitudes into ratios and then taking
a simple inverse. I'm surprised you got past primary
school on this evidence!

George

On 23 Sep, 21:15, HW@....(Dr. Henri Wilson) wrote:
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.

It is not a theory of any kind Henry. A theory
consists of a set of eqations that make
quanititative predictions of observed measurable
values.

What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

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.

What I have written above are the facts, it is
up to you to interpret them. Photons only ever
cause a single detection in a sensor (perhaps
multiple electrons if avalanche technology is
used) and individual photons are subject to
all the normalrules of interference and
diffraction.

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?

I explained that before. Photons(idividually) from
one side of the star create a pattern of intensity.
Photons from the other also create their own pattern,
again behaving individually, but the patterns are
displaced by a small amount, essentially the size
of the "image" formed by the telescope. Because one
is shifted relative to the other, the nulls don't
occur in the same place which affects the contrast
ratio of the fringe pattern. The actual analysis is
much more complex, involving limb darkening and lots
of other stuff but those are the basics.

snip lots not commented

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.

Of course not, as I told you before, it is the
superposition of the displaced interference
patterns that carries the information.

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.

Tough, the radius is observed to vary and the
interferometric, photometric and integrated
velocity measurements are all similar.

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. ...

Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

George

On 23 Sep, 22:18, HW@....(Dr. Henri Wilson) wrote:
On Fri, 21 Sep 2007 15:46:12 +0100, "George Dishman" wrote:

much snipped as not addressed

....
.. 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.

Yes of course... but outward energy flow has to be consistent...

Of course, typically the flow is analysed along a
column.

I once had a job taking very fine photos of the sun, mainly in the H red.

The edges of the ball were quite sharp.

First remember the 400km I quoted is vertically into
the Sun at the centre. As you move towards the limb
it becomes smaller as the light path is at a shallower
angle. At this range even 400km would subtend a very
small angle at the camera.

However, that doesn't mean a significant thick tansparent layer does not
exist....

Of course, the corona for example.

and that layer emits plenty of light.

Oops, no. You forgot Kirchofff's Law again, if it is
really transparent it cannot emit at all. In practice
both the outer layer just above the photosphere and
the corona show spectral lines. The upper surface is
cooler than the photosphere so they appear in
absorption while the corona is hotter and the lines
show as emission. What we have been discussing re
temperature is the background continuum excluding the
lines.

George

On Tue, 25 Sep 2007 05:47:38 -0700, George Dishman
wrote:

On 23 Sep, 21:15, HW@....(Dr. Henri Wilson) wrote:
On Wed, 19 Sep 2007 22:16:11 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...



...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.

It is not a theory of any kind Henry. A theory
consists of a set of eqations that make
quanititative predictions of observed measurable
values.

What I said is a statement of what is observed,
any valid theory has to predict that behaviour.

George, admit it. Nobody including you has much of a clue about the true nature
of light.


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.

What I have written above are the facts, it is
up to you to interpret them. Photons only ever
cause a single detection in a sensor (perhaps
multiple electrons if avalanche technology is
used) and individual photons are subject to
all the normalrules of interference and
diffraction.

To get interference from two detectors spaced 100 metres apart requires
coherence. If a single photon stretches right across then what you said above
is wrong. A 100 m wide photon will strike more than one atom.
Even YOU should see that.



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?

I explained that before. Photons(idividually) from
one side of the star create a pattern of intensity.
Photons from the other also create their own pattern,
again behaving individually, but the patterns are
displaced by a small amount, essentially the size
of the "image" formed by the telescope. Because one
is shifted relative to the other, the nulls don't
occur in the same place which affects the contrast
ratio of the fringe pattern. The actual analysis is
much more complex, involving limb darkening and lots
of other stuff but those are the basics.

It sounds as though all you are seeing is a fuzzy image of the star.


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.

Of course not, as I told you before, it is the
superposition of the displaced interference
patterns that carries the information....

I could just as easily carry information about varying light speed.


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.

Tough, the radius is observed to vary and the
interferometric, photometric and integrated
velocity measurements are all similar.

..they are willusions.

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. ...

Don't waste your time with stupid word games, it would
be better spent if you learned some schoolboy calculus.

I know more about calculus than you do. I prefer to avoid it where possible.

George

Henri Wilson. ASTC,BSc,DSc(T)

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

On Tue, 25 Sep 2007 05:57:48 -0700, George Dishman
wrote:

On 23 Sep, 22:18, HW@....(Dr. Henri Wilson) wrote:
On Fri, 21 Sep 2007 15:46:12 +0100, "George Dishman" wrote:

much snipped as not addressed

...
.. 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.

Yes of course... but outward energy flow has to be consistent...

Of course, typically the flow is analysed along a
column.

I once had a job taking very fine photos of the sun, mainly in the H red.

The edges of the ball were quite sharp.

First remember the 400km I quoted is vertically into
the Sun at the centre. As you move towards the limb
it becomes smaller as the light path is at a shallower
angle. At this range even 400km would subtend a very
small angle at the camera.

However, that doesn't mean a significant thick tansparent layer does not
exist....

Of course, the corona for example.

and that layer emits plenty of light.

Oops, no. You forgot Kirchofff's Law again, if it is
really transparent it cannot emit at all.

George, absorption is exponential.
If the layer absorbs only 10% per 100 kms, about 36% will still pass through
1000 kms. The layer will have an emissivity 0 and will still radiate.

In practice
both the outer layer just above the photosphere and
the corona show spectral lines. The upper surface is
cooler than the photosphere so they appear in
absorption while the corona is hotter and the lines
show as emission. What we have been discussing re
temperature is the background continuum excluding the
lines.

....yes..and I have been pointing out that ADoppler can cause a considerable
shift in the planck curve.


George

Henri Wilson. ASTC,BSc,DSc(T)

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

On 25 Sep, 22:53, HW@....(Dr. Henri Wilson) wrote:
On Tue, 25 Sep 2007 05:57:48 -0700, George Dishman wrote:
On 23 Sep, 22:18, HW@....(Dr. Henri Wilson) wrote:
On Fri, 21 Sep 2007 15:46:12 +0100, "George Dishman" wrote:

much snipped as not addressed

...
.. 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.

Yes of course... but outward energy flow has to be consistent...

Of course, typically the flow is analysed along a
column.

I once had a job taking very fine photos of the sun, mainly in the H red.

The edges of the ball were quite sharp.

First remember the 400km I quoted is vertically into
the Sun at the centre. As you move towards the limb
it becomes smaller as the light path is at a shallower
angle. At this range even 400km would subtend a very
small angle at the camera.

However, that doesn't mean a significant thick tansparent layer does not
exist....

Of course, the corona for example.

and that layer emits plenty of light.

Oops, no. You forgot Kirchofff's Law again, if it is
really transparent it cannot emit at all.

George, absorption is exponential.
If the layer absorbs only 10% per 100 kms, about 36% will still pass through
1000 kms. The layer will have an emissivity 0 and will still radiate.

See the next paragraph:

In practice
both the outer layer just above the photosphere and
the corona show spectral lines. The upper surface is
cooler than the photosphere so they appear in
absorption while the corona is hotter and the lines
show as emission. What we have been discussing re
temperature is the background continuum excluding the
lines.

There is no continuum emission from the 'transparent'
layer but where there is a resonance and the opacity
is higher at that specific frequency, we see absorption.

...yes..and I have been pointing out that ADoppler can cause a considerable
shift in the planck curve.

And I have been pointing that while it _could_, we
know it _doesn't_, the actual shift is only 0.01%
and such a small shift doesn't affect temperature
determination.

Note also that if the frequency shift were larger,
it would no longer be a black body curve but in
practice the shift is so small that is negligible.

George