Why are the 'Fixed Stars' so FIXED?

Jerry wrote:

This is the way Wilson's theory ends
This is the way Wilson's theory ends
This is the way Wilson's theory ends
Not with a bang but a whimper.

From "The Hollow Theory"?

We are the hollow theories
We are the crackpot theories
etc.

--
Thomas M. Sommers -- -- AB2SB

On Aug 14, 12:15 pm, "T.M. Sommers" wrote:
Jerry wrote:

This is the way Wilson's theory ends
This is the way Wilson's theory ends
This is the way Wilson's theory ends
Not with a bang but a whimper.

From "The Hollow Theory"?

We are the hollow theories
We are the crackpot theories
etc.

--
Thomas M. Sommers

OK, OK, I -admit- to plagiarizing your poem
"The Hollow Theories", by Thomas Sommers Eliot (T. S. Eliot)
I thought it would be fun to substitute "Wilson's" for "crank"
in your original verse, "This is the way crank theory ends"

You won't sue, I hope? (shudder...)

Jerry

On Tue, 14 Aug 2007 05:10:47 -0700, George Dishman
wrote:


Henri Wilson wrote:

On Tue, 14 Aug 2007 00:26:32 -0700, George Dishman wrote:
Henri Wilson wrote:
Not even close, most of the V band is the temperature
effect, note that the K band and the top velocity curve
are not very similar at all.

Well if that is the case then there is no problem.

To produce the V band curve, I require e=0.21 and yaw = -47
For the K band the figures are e= 0.15 and yaw = -12

Right, two different sets of parameters for the same source.

George, don't feign stupidity. Remember you're a senior engineer...

It is NOT the same source. On average, the bands come from different layers.

Sorry Henry, it is not separate layers, the gas emits black body
radiation but since we are talking about the light that reaches
us, obviously that nearest us must be nearly transparent. There
isn't a sharp cutoff but a rapid increase in opacity with depth
matched by a slight increase in temperature. We see all the
frequencies from all the range of depth.

I said, "on average".
I stick by my statement.

program shows that...but you wont rad about it in any published paper.

No it doesn't, all you have done is match a Keplerian orbit
model to a black body curve at each of two temperatures,
your program tells you nothing at all because it is modelling
the wrong thing entirely.

the use of Keplarian analogy clearly works and is therefore perfectly
legitimate. It is also very revealing. the example above is a good example of
how new facts can emerge using BaTh.

As I pointed out before, the eccentricity signifies the variation in the
'springiness' of hte expansion/contraction whilst 'yaw angle' relates to a kind
of hysteresis in the loop.

This merely suggests that the layers emitting these bands are not quite the
same on average ...

There aren't separate layers emitting different bands Henry,
the emission is almost thermal with contributions across
all bands from throughout the depth of a quite thin layer
(400km for the Sun).

George, they don't have to be very different

Henry, the material is not frequency selective, it is
black body at any given depth.

George, quite obviously there is a radial temperature gradient...sufficient to
cause the two bands to originate ON AVERAGE from slightly diffrent layers.

I think it is time you became a bit more positive George.

and that one is exactly not following the other in either phase
or radial velocity change.

There is a single layer involved in the emission process.

No, I have shown otherwise.

No, you have simply misunderstood what you are seeing.
The difference in the bands fits the Planck curve integrated
over the depth at any time in the cycle, but the mean
temperature of the layer changes with the pusation phase.

Te light comes from a fairly thin layer that has a temperature gradient.
What the hell is so amazing about that?

Of
course the driving force for the instability is another layer
somewhat deeper into the star but we cannot see that
directly as the intervening material is opaque. That layer is
at about 40,000K and 5% of the radius in from the photosphere
which is at about 6,000K. It is that outer layer whose motion
you are trying to describe.

George, the whole astronomy world is about to be turned on its head as I
prepare to publish what will be the biggest bombshell in the history of
physics.

Yet more crank garbage Henry? Everything you have claimed
has been disproven over the last few weeks by your own
program so nothing you might write will have any impact
whatsoever. Still it might rate an entry or two in the "Bad
Astronomy" site.

You can start burning your books now George....My theory has survived every bit
of criticism you have thrown ai it. It is alos producing some amazin
discoveries already.



EINSTEIN WAS WRONG.
Accept it!

Einstein's work matches what is observed, Ritz's doesn't.

Ritz died before he could finish the theory.

When you get round to modelling L Car properly, you will
discover it would require that any effects must be VDoppler
only, just as we found for the pulsars and contact binaries.

The astronomical evidence proves Einstein right every time,
even using your own program. Think back to the best test
we have found, pulsar J1909-3744 where the pulses are
within 74ns of the arrival time predicted by GR _including_
the Shapiro delay which ballistic theory gets completely
wrong.

Waving your hands and pretending your results are something
other than what you actually got won't salvage your nonsense,
and shouting at people to accept your failed theory won't
change their minds either, it only makes your desparation
apparent.

Give up George. Accept the obvious....you have been backing a loser on purely
religious grounds.

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 Tue, 14 Aug 2007 05:54:00 -0700, Jerry
wrote:


On Aug 14, 4:50 am, HW@....(Henri Wilson) wrote:
On Tue, 14 Aug 2007 00:26:32 -0700, George Dishman
wrote:
Henri Wilson wrote:

Not even close, most of the V band is the temperature
effect, note that the K band and the top velocity curve
are not very similar at all.

Well if that is the case then there is no problem.

To produce the V band curve, I require e=0.21 and yaw = -47
For the K band the figures are e= 0.15 and yaw = -12

Right, two different sets of parameters for the same source.

George, don't feign stupidity. Remember you're a senior engineer...

It is NOT the same source. On average, the bands come from
different layers. my program shows that...but you wont rad
about it in any published paper.

Henri, the number of self-contradictory concepts that you need
to believe in for your theory to work is staggering.

How deeply separated are these hypothetical layers of yours? I
can't say exactly, but here is a rough estimate: Multiple
measurement methods converge on an angular diameter for L Car
of approximately 3 mas. The Hipparcos parallax is 7.59 mas, so
the distance is about 132 parsec. The radius of L Car is hence
approximately 30,000,000 km. The ratios of the semimajor and
semiminor axes of the two ellipses are 1.53 and 1.35. Taking
the difference and multiplying by the radius tells us that the
separation between layers must be at least 5,000,000 km or the
layers overlap at some point in their pulsation cycle.

You are claiming, therefore, that the apparent diameter of the
star may varies something on the order of 18 percent depending
on whether one is observing in V band or K band, that light from
5,000,000 km within the star does not suffer rapid extinction
as it travels through dense layers of stellar material, and that
everything known about black body radiation is false.

....theories, theories...all based wrongly on constant c.

As I pointed out before, the eccentricity signifies the
variation in the 'springiness' of hte expansion/contraction
whilst 'yaw angle' relates to a kind of hysteresis in the loop.

This merely suggests that the layers emitting these bands are
not quite the same on average ...

There aren't separate layers emitting different bands Henry,
the emission is almost thermal with contributions across
all bands from throughout the depth of a quite thin layer
(400km for the Sun).

George, they don't have to be very different

and that one is exactly not following the other in either
phase or radial velocity change.

There is a single layer involved in the emission process.

No, I have shown otherwise.

Utter fantasy.


Of
course the driving force for the instability is another layer
somewhat deeper into the star but we cannot see that
directly as the intervening material is opaque. That layer is
at about 40,000K and 5% of the radius in from the photosphere
which is at about 6,000K. It is that outer layer whose motion
you are trying to describe.

George, the whole astronomy world is about to be turned on its
head as I prepare to publish what will be the biggest bombshell
in the history of physics.

This is the way Wilson's theory ends
This is the way Wilson's theory ends
This is the way Wilson's theory ends
Not with a bang but a whimper.

EINSTEIN WAS WRONG.
Accept it!

Jerry

Read my reply to 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 Aug 14, 7:59 pm, HW@....(Henri Wilson) wrote:
On Tue, 14 Aug 2007 05:54:00 -0700, Jerry
wrote:
On Aug 14, 4:50 am, HW@....(Henri Wilson) wrote:
On Tue, 14 Aug 2007 00:26:32 -0700, George Dishman
wrote:

Right, two different sets of parameters for the same source.

George, don't feign stupidity. Remember you're a senior engineer...

It is NOT the same source. On average, the bands come from
different layers. my program shows that...but you wont rad
about it in any published paper.

Henri, the number of self-contradictory concepts that you need
to believe in for your theory to work is staggering.

How deeply separated are these hypothetical layers of yours? I
can't say exactly, but here is a rough estimate: Multiple
measurement methods converge on an angular diameter for L Car
of approximately 3 mas. The Hipparcos parallax is 7.59 mas, so
the distance is about 132 parsec. The radius of L Car is hence
approximately 30,000,000 km. The ratios of the semimajor and
semiminor axes of the two ellipses are 1.53 and 1.35. Taking
the difference and multiplying by the radius tells us that the
separation between layers must be at least 5,000,000 km or the
layers overlap at some point in their pulsation cycle.

You are claiming, therefore, that the apparent diameter of the
star may varies something on the order of 18 percent depending
on whether one is observing in V band or K band, that light from
5,000,000 km within the star does not suffer rapid extinction
as it travels through dense layers of stellar material, and that
everything known about black body radiation is false.

...theories, theories...all based wrongly on constant c.

Nope. Assumption of constant c is not required at all.

Remember, multiple independent measurement methods yield the same
angular diameter for L Car. Consider the method based on luminance.

The Planck radiation law tells us that the intensity of emission
at a given frequency f is given by
I(f) = (2hf^3/c^2)(1/(exp(hf/(kT))-1) [Eq 1]
When kT hf, exp(hf/(kT)) ~= 1+hf/(kT) so that the above
simplifies to
I(f) = 2f^2*kT/c^2 [Eq 2]

In other words, the observed intensity in a given narrow band of
frequencies is APPROXIMATELY proportional to the absolute temp.
The lower the frequency and/or the higher the temperature, the
better the approximation. (The approximation is mediocre for
V-band light, so we expect some discrepancy between results
obtained using this approximation versus results obtained using
a more accurate method.)

No speed of light assumptions so far!

L Car has spectral type G5 at maximum, hence a temperature about
5500K. The Sun has spectral type G2, hence a temperature about
5800K. Therefore, we conclude that on an area-per-area basis,
L Carinae is about 94.8% as bright as the Sun in wavelength
ranges where the approximation in Equation 2 is valid.

The visual magnitude of the Sun is -26.73
The angular diameter of the Sun is 31'28" = 1888 arc seconds

The visual magnitude of L Carinae is approximately 3.30 at
visual maximum, which occurs near minimum radius.

Q: What is the angular diameter of L Carinae?
A: The Sun appears 1.03e12 times brighter than L Carinae, hence
spans an apparent area 1.03e12*0.948 = 9.75e11 times the area
of L Car. The square root is 9.87e5. Divide 1888 by 9.87e5 to
get 1.9 mas at visual maximum, which is reasonably close to the
published figures of 2.5 to 2.7 mas which were made using more
accurate assumptions and fewer approximations.

The Hipparcos parallax is 7.59 mas, so the distance is about
132 parsec. Even YOU should be able to figure that out.

Combine these pieces of information, and you can't escape the
conclusion that YOUR THEORY says that light travels from millions
of km beneath the surface of L Carinae WITHOUT EXTINCTION to
emerge at a velocity c+v which is distinct from c+v' light emitted
from the photosphere...

....which of course is absolute nonsense, even according to
YOUR THEORY...

There aren't separate layers emitting different bands Henry,
the emission is almost thermal with contributions across
all bands from throughout the depth of a quite thin layer
(400km for the Sun).

George, they don't have to be very different

Oh, yes they do, Henri. Yes they do...

There is a single layer involved in the emission process.

No, I have shown otherwise.

Utter fantasy.

Of
course the driving force for the instability is another layer
somewhat deeper into the star but we cannot see that
directly as the intervening material is opaque. That layer is
at about 40,000K and 5% of the radius in from the photosphere
which is at about 6,000K. It is that outer layer whose motion
you are trying to describe.

George, the whole astronomy world is about to be turned on its
head as I prepare to publish what will be the biggest bombshell
in the history of physics.

This is the way Wilson's theory ends
This is the way Wilson's theory ends
This is the way Wilson's theory ends
Not with a bang but a whimper.

EINSTEIN WAS WRONG.
Accept it!

Jerry

Read my reply to George.

Jerry
Henri Wilson's Lies
(1)Fakes Diploma (2)Uses Deceptive Language (3)Fakes Program
(4)Intentionally Misquotes (5)Snips (6)Accuses Others of Lying
1 http://mysite.verizon.net/cephalobus...ri/diploma.htm
2 http://mysite.verizon.net/cephalobus.../deception.htm
3 http://mysite.verizon.net/cephalobus...rt_aurigae.htm
4 http://mysite.verizon.net/cephalobus...ri/history.htm
5 http://mysite.verizon.net/cephalobus...enri/snips.htm
6 http://mysite.verizon.net/cephalobus...ri/accuses.htm

Jerry ) writes:
On Aug 14, 12:15 pm, "T.M. Sommers" wrote:
Jerry wrote:

This is the way Wilson's theory ends
This is the way Wilson's theory ends
This is the way Wilson's theory ends
Not with a bang but a whimper.

From "The Hollow Theory"?

We are the hollow theories
We are the crackpot theories
etc.

--
Thomas M. Sommers

OK, OK, I -admit- to plagiarizing your poem
"The Hollow Theories", by Thomas Sommers Eliot (T. S. Eliot)
I thought it would be fun to substitute "Wilson's" for "crank"
in your original verse, "This is the way crank theory ends"


What's the one that starts

Astro is cruellest group
Breeding lunacies out of the dead minds ?

--John Park

Henri Wilson wrote:

On Tue, 14 Aug 2007 05:10:47 -0700, George Dishman wrote:
Henri Wilson wrote:
On Tue, 14 Aug 2007 00:26:32 -0700, George Dishman wrote:
Henri Wilson wrote:
Not even close, most of the V band is the temperature
effect, note that the K band and the top velocity curve
are not very similar at all.

Well if that is the case then there is no problem.

To produce the V band curve, I require e=0.21 and yaw = -47
For the K band the figures are e= 0.15 and yaw = -12

Right, two different sets of parameters for the same source.

George, don't feign stupidity. Remember you're a senior engineer...

It is NOT the same source. On average, the bands come from different layers.

Sorry Henry, it is not separate layers, the gas emits black body
radiation but since we are talking about the light that reaches
us, obviously that nearest us must be nearly transparent. There
isn't a sharp cutoff but a rapid increase in opacity with depth
matched by a slight increase in temperature. We see all the
frequencies from all the range of depth.

I said, "on average".

And I said the curve was the weighted integral over the
depth, essentially we are saying the same.

I stick by my statement.

Later you change to agree with my view, I think it is more
one of emphasis.

program shows that...but you wont rad about it in any published paper.

No it doesn't, all you have done is match a Keplerian orbit
model to a black body curve at each of two temperatures,
your program tells you nothing at all because it is modelling
the wrong thing entirely.

the use of Keplarian analogy clearly works and is therefore perfectly
legitimate. It is also very revealing. the example above is a good example of
how new facts can emerge using BaTh.

Except that it isn't a new fact at all, conventional theory
alreay provides the relative contribution as a function
of depth and tells us the thickness which very thin in
the context.

As I pointed out before, the eccentricity signifies the variation in the
'springiness' of hte expansion/contraction whilst 'yaw angle' relates to a kind
of hysteresis in the loop.

This merely suggests that the layers emitting these bands are not quite the
same on average ...

There aren't separate layers emitting different bands Henry,
the emission is almost thermal with contributions across
all bands from throughout the depth of a quite thin layer
(400km for the Sun).

George, they don't have to be very different

Henry, the material is not frequency selective, it is
black body at any given depth.

George, quite obviously there is a radial temperature gradient...

Yes, obviously.

sufficient to
cause the two bands to originate ON AVERAGE from slightly diffrent layers.

No, the difference is small compared to the thickness,
both bands integrate over the same region.

I think it is time you became a bit more positive George.

I think it is time you did some research and found out
the facts instead of waving your hands and hoping for
the best.

and that one is exactly not following the other in either phase
or radial velocity change.

There is a single layer involved in the emission process.

No, I have shown otherwise.

No, you have simply misunderstood what you are seeing.
The difference in the bands fits the Planck curve integrated
over the depth at any time in the cycle, but the mean
temperature of the layer changes with the pusation phase.

Te light comes from a fairly thin layer that has a temperature gradient.
What the hell is so amazing about that?

Nothing at all, and both bands receive the integrated total of
light across that depth, what the hell is so amazing about
that?

Of
course the driving force for the instability is another layer
somewhat deeper into the star but we cannot see that
directly as the intervening material is opaque. That layer is
at about 40,000K and 5% of the radius in from the photosphere
which is at about 6,000K. It is that outer layer whose motion
you are trying to describe.

George, the whole astronomy world is about to be turned on its head as I
prepare to publish what will be the biggest bombshell in the history of
physics.

Yet more crank garbage Henry? Everything you have claimed
has been disproven over the last few weeks by your own
program so nothing you might write will have any impact
whatsoever. Still it might rate an entry or two in the "Bad
Astronomy" site.

You can start burning your books now George....My theory has survived every bit
of criticism you have thrown ai it.

ROFL, it cannot even predict a fringe shift for the Sagnac
experiment. Your two current equations predict a null result
forSagnac, predict an advance instead of a delay for the
Shapiro effect, predict positional errors of tens of metres in
GPS, and require a "speed equalisation" distance so short
for pulsars that it reduces to a copy of aether theory with
no ballistic efefcts detectable. It hasn't survived a single test.

It is alos producing some amazin
discoveries already.

Sure, it's great at generating fantasies.

EINSTEIN WAS WRONG.
Accept it!

Einstein's work matches what is observed, Ritz's doesn't.

Ritz died before he could finish the theory.

As you said, I stick by my statement. Your single addition
doesn't solve any of the problems apart from the pulsars
where it reduces the result to the same as SR.

When you get round to modelling L Car properly, you will
discover it would require that any effects must be VDoppler
only, just as we found for the pulsars and contact binaries.

The astronomical evidence proves Einstein right every time,
even using your own program. Think back to the best test
we have found, pulsar J1909-3744 where the pulses are
within 74ns of the arrival time predicted by GR _including_
the Shapiro delay which ballistic theory gets completely
wrong.

Waving your hands and pretending your results are something
other than what you actually got won't salvage your nonsense,
and shouting at people to accept your failed theory won't
change their minds either, it only makes your desparation
apparent.

Give up George. Accept the obvious....you have been backing a loser on purely
religious grounds.

I accept the obvious, Ritz's theory gives wrong predictions for
every known test other than the MMX, it is and always was a
dead duck. GR, SR and QED on the other hand get every test
right. Those are the facts.

George

John Park wrote:
Jerry ) writes:
On Aug 14, 12:15 pm, "T.M. Sommers" wrote:
Jerry wrote:

This is the way Wilson's theory ends
This is the way Wilson's theory ends
This is the way Wilson's theory ends
Not with a bang but a whimper.

From "The Hollow Theory"?

We are the hollow theories
We are the crackpot theories
etc.

OK, OK, I -admit- to plagiarizing your poem
"The Hollow Theories", by Thomas Sommers Eliot (T. S. Eliot)
I thought it would be fun to substitute "Wilson's" for "crank"
in your original verse, "This is the way crank theory ends"

What's the one that starts

Astro is cruellest group
Breeding lunacies out of the dead minds ?

Wasn't it "The Love Song of J. Albert Einstein"?

--
Thomas M. Sommers -- -- AB2SB

Jerry wrote:
On Aug 14, 12:15 pm, "T.M. Sommers" wrote:
Jerry wrote:

This is the way Wilson's theory ends
This is the way Wilson's theory ends
This is the way Wilson's theory ends
Not with a bang but a whimper.

From "The Hollow Theory"?

We are the hollow theories
We are the crackpot theories
etc.

OK, OK, I -admit- to plagiarizing your poem
"The Hollow Theories", by Thomas Sommers Eliot (T. S. Eliot)
I thought it would be fun to substitute "Wilson's" for "crank"
in your original verse, "This is the way crank theory ends"

You won't sue, I hope? (shudder...)

Sssh! I don't want TSE to sue me.

--
Thomas M. Sommers -- -- AB2SB

"Henri Wilson" HW@.... wrote in message
...
On Sun, 12 Aug 2007 16:03:31 +0100, "George Dishman"

wrote:


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

ADoppler can easily produce 15x or more before peaking..

Sure, but if you then say "If the brightness curve
is ADoppler, then so will be the velocity curve."
you then get a prediction observed speeds close to
the speed of light instead of the tens of km/s we
actually see.

'K', George, 'K'.

K=1 Henry, or if this Cepheid is atypical it should
have a velocity curve peaking at thousands of km/s.

Go away George. You are just trolling, now.

No Henry, just reminding you of why K doesn't
solve the problem even if it didn't make the
theory self-contradictory. You need K=0 for
some observations but K=1 for others and for
Cepheids you need both at the same time.

Put simply, photon bunching continues long after individual photon
shrinking
virtually ceases.

Not in this case, unless the velocity peak is 90 degrees
out from the luminosity.

George this is obviously far too hard for you.

More petty attepts at insults? That means you
kinow I'm right and have no answer to the point,
you are stalling for time again.

The main point is that the willusory velocity and brightness curves will
be
generally in phase or close to it.

Right, so the velocity peaks at the time of your
so-called eclipse - wrong!

No George, The ACCELERATION peaks.

And if the curves are in phase then so does the
velocity according to you, something that is
obviously impossible.

Nah, it's trivial stuff, you just need to apply
it consistently and stop trying to hide when it
gives you a problem.

You can't do it...I didn't think so....

http://www.georgedishman.f2s.com/Henri/sine_right.png

All you did was take a curve and differentiate it twice.

Nope, I added the 'willusion' effect too.

when are you going to produce a curve George.

That's it.




George, I know I shouldn't be telling an 'expert' he is wrong but
standing
waves will only occur if the 'wavecrest arrival frequency' is an exact
multiple
of the speaker frequency.

Nope, this is an entirely new phenomenon you haven't
even considered. You will need to lay aside
preconceptions and work it through from first
principles.

You'll just get noise.

Nope, you get standing waves. This is obviously
too hard for you ;-)

So naive and revealing that it gets the right answers every time....

What does that suggest George..

That you picked the answer out of thin air because it
suits you without doing the work to see if your guess
was derivable from your theory. It isn't, the results
will surprise you.

stop trolling George.

Do the maths, you wil find I am correct.

Still me, it is _less_ massive than Jupiter.

Well i didn't check the report so i don't know.

I did, I do, I have highlighted it above.

It isn't important anyway.

No, just another example of how you quote stuff
incorrectly because you only see what you want
to see, even if it isn't there.

....
Yet again you lie. As I gave you befo

http://arxiv.org/abs/astro-ph/0402244

Fig 1, top panel. Here is my fit:

http://www.georgedishman.f2s.com/Henri/fit_vel.png

and my prediction for the radius curve of fig. 3:

http://www.georgedishman.f2s.com/Henri/fit_rad.png

Now where are yours?

The V band velocity and rmagnitude curves are very similar. That's what
I
will produce, too. This is a simple exercise.

Note that the K band sets the maximum ballistic
effect, the additional effect in V must be due
to temperature and that is at least 60% of the
variation. Any fits you do to that will be
worthless.

George, I know it must be very painful for you to see the BaTh having one
success after another.

ROFL, it has failed _every_ single test we have
looked at!

George