Why are the 'Fixed Stars' so FIXED?

On Sat, 01 Sep 2007 03:31:04 +0200, "Paul B. Andersen"
wrote:

Androcles wrote:
"Paul B. Andersen" wrote in message
...
: Henri Wilson wrote:
: Explain the 90 deg phase lag then George.
:
: Not hard at all to explain why the curves are different.
: If the radius of the star didn't change, it is obvious
: from Planck's blackbody equation that the luminosity
: variation due to the changing temperature is much
: bigger in visible light (V-band) than it is in IR (K-band 2.2u).

Everything is "obvious". Obviously you are a lunatic.

I note with a yawn that Androcles doesn't find the obvious
consequence of Planck's black body radiation law to be obvious.

Nothing is obvious in a haze, is it?

Androcles is an undiscovered genius....leave him alone..

Paul



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

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

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

Henri Wilson wrote:
On Sun, 26 Aug 2007 23:14:54 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
Explain the 90 deg phase lag then George.
Not hard at all to explain why the curves are different.
If the radius of the star didn't change, it is obvious
from Planck's blackbody equation that the luminosity
variation due to the changing temperature is much
bigger in visible light (V-band) than it is in IR (K-band 2.2u).
So since the radius changes as well, it is clear that
the luminosity variation due to the changing surface area
will be relatively more important in IR than it is in visible light.
That's why the IR-light curve has it maximum and minimum
at ca. phase 0.4 and 0.9 respectivly, just like the angular
diameter curve.
Compare fig 2 (K-mag) and fig2 and 3.
http://www.arxiv.org/PS_cache/astro-.../0402244v1.pdf

The V light curve will be more dominated by the temperature,
which has its maximum and minimum at phase 1 and 0.65 respectively.
Compare fig 2. (V-mag) in document above to fig 4.7 in document
below:
http://ses.library.usyd.edu.au/bitst....0014whole.pdf
http://tinyurl.com/26q3xh
...and I'm sure you could produce an equally nonsensical theory if the
published figures were entirely diffent.

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

It is reasonably well confirmed.

It is extremely well confirmed.
So why did you call it nonsensical?

What isn't confirmed is whether or not the
average cepheid has a black body spectrum. Nor has it been confirmed that its
spectrum would remain black body if it went 'huff puff' all day long.

As my calculation shows, the observed light curves in K and V from l Carinae
are consistent with a black body spectrum.

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

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

That's a bloody PhD thesis....Do you really think a Phd student is going to
stand up and announce that the whole of astronomy is bull**** because it is
based on all starlight traveling to little planet Earth at precisely speed 'c'?
Of course not Tussellad...

In other words, you won't expect a Phd student to be a crank. :-)
You have to utterly ignorant of physics and astronomy to be open minded
enough to claim that "the whole of astronomy is bull****", right? :-)

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

....so you are already on the wrong track....

Hardly a convincing argumentation for why the temperature curve
in fig. 4.3 must be wrong. :-)

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

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

Interferometry relies on all starlight moving at constant c wrt Earth.
Since that is not true, the technique can best be regarded as highly suspect.
The published cuvre is lousy anyway...nothing like a best fit....

Don't be such a blatant idiot, Henri.
If the BaTh predicts that it is possible to use a telescope
to measure the angular distance between two stars, then it also predicts
that it is possible to use an interferometer to measure the angular
diameter of a star. This is simple geometry, Henri. The geometry
is the same in both cases, and it doesn't matter what the speed
of light is.

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

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

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

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

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

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

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

It is highly possible that some stars DO pulsate.

I take this remark as an admission that you accept
that l Carinae is a pulsating star, and the reason for
why the K and V light curves are as they are simply is
Planck's law.

So why did you call this explanation 'nonsensical'?

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

Do you still find Planck's blackbody radiation law nonsensical, Henri?

Planck's law was empirically derived for what is assumed to be a perfect black
body.

Planck's law can be derived from the assumption (postulate) that
the energy of the oscillators in the radiating body is quantized.

Stars vary considerably and not many fit that curve well at all.

Nonsense.
All stellar spectra are black body spectra with absorption and
emission lines due to the fact that the black body radiation
from the stellar photosphere has to go through the higher layers
of the stellar atmosphere.
The spectrum is given primarily by the surface temperature of
the star, and varies remarkably little between stars with equal
temperatures.

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

But you can of course claim the right to believe otherwise
according to the UN freedom of religion act. :-)

Paul

In sci.physics.relativity, Androcles

wrote
on Sun, 02 Sep 2007 11:10:04 GMT
:

"The Ghost In The Machine" wrote in message
...
: In sci.physics.relativity, Androcles
:
: wrote
: on Sat, 01 Sep 2007 14:03:45 GMT
: :
:
: "Paul B. Andersen" wrote in message
: ...
: : Androcles wrote:
: : "Paul B. Andersen" wrote in message
: : ...
: : : Androcles wrote:
: : : "Paul B. Andersen" wrote in
: message
: : : ...
: : : : Henri Wilson wrote:
: : : : Explain the 90 deg phase lag then George.
: : : :
: : : : Not hard at all to explain why the curves are different.
: : : : If the radius of the star didn't change, it is obvious
: : : : from Planck's blackbody equation that the luminosity
: : : : variation due to the changing temperature is much
: : : : bigger in visible light (V-band) than it is in IR (K-band
2.2u).
: : :
: : : Everything is "obvious". Obviously you are a lunatic.
: : :
: : : I note with a yawn that Androcles doesn't find the obvious
: : : consequence of Planck's black body radiation law to be obvious.
: : :
: : : Nothing is obvious in a haze, is it?
: :
: : ASSistant Professor "Paul B. Andersen" of :
: : Agder University College (HiA)
: : Serviceboks 422, N-4604 Kristiansand, NORWAY Tel (+47) 38 14 10 00
Fax
: : (+47) 38 14 10 01
: : has executed the biggest fumble ever seen in the history of
: : sci.physics.relativity
: : in message
: : ...
: :
: : "The spectral class [of stars] is determined by the relative
positions
: : and intensities
: : of the absorption lines, and these are unaffected by a Doppler
shift."
: :
: : The all time classic:
: :
: : "That is, we can reverse the directions of the frames
: : which is the same as interchanging the frames,
: : which - as I have told you a LOT of times,
: : OBVIOUSLY will lead to the transform:
: : t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
: : x = (xi - v*tau)/sqrt(1-v^2/c^2)
: : or:
: : tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
: : xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
: :
: :
: The faster you go the longer it takes to get there, OBVIOUSLY.
: yawn
:
: Here's how I look at it.

[snip wrong argument]

... reach the star, d-vt = 0.

The Andersen Transforms are d+vt 0.
The faster you go the longer it takes to get there, OBVIOUSLY.
yawn

Since I was assuming a different transform:

xi = (x - vt) * g
tau = (t - vx/c^2) * g

this doesn't quite work. However, I'd have to dig through
the posts to see what Paul was claiming and your objections
thereto, and I'd highly prefer a different nomenclature anyway,
something along the following lines.

If E is a symbol for the coordinate frame for Earth and S
the symbol for the coordinate frame for the spacecraft,
and the spacecraft is traveling at a uniform velocity
v away from Earth in a straight line, and with
coincident origins, one can establish the usual Lorentz transform

x_S = (x_E - v * t_E) / sqrt(1-v^2/c^2)
t_S = (t_E - v * x_E / c^2) / sqrt(1-v^2/c^2)

This transform is invertable, with a little work,
resulting in:

x_E = (x_S + v * t_S) / sqrt(1-v^2/c^2)
t_E = (t_S + v * x_S / c^2) / sqrt(1-v^2/c^2)

I can also express these using the notation

(x_E,t_E)_E = ( (x_E - vt_E)*g, (t_E - v*x_E/c^2)*g)_S
(x_S,t_S)_S = ( (x_S + vt_S)*g, (t_S + v*x_S/c^2)*g)_E

where _S and _E establish which frame I'm using for the
given coordinate point, and g = 1/sqrt(1-v^2/c^2) the
usual gamma. I'm not sure if this is a proper tensor,
but it does establish a mapping from E-frame to S-frame
(and vice versa; the transform is bijective for any v in
the open interval (-1, +1)), and I for one think it helps
clarify things, as one doesn't have to ask such questions
as "which frame is this relevant in?".

Now one can set up various problems: the most obvious
one is when does SR predict that the spacecraft reaches
the star? More specifically, what values of x_E, t_E,
x_S, and t_S are such that
(x_E, t_E)_E = (d,t_E)_E [Earth sees it reaching the star][*]
and
(x_S, t_S)_S = (0,t_S)_S ? [The spacecraft sees itself reaching the star]

The first equality should be obvious, but the second one might not be.
The general issue is that an astronaut cannot leave his ship, and since
the ship defines the ship-frame, x_S = 0 for most of the interesting
measurements, unless one postulates the astronaut holding a rod of
a certain length (which is a slightly different problem).

So now we know x_E = d, x_S = 0. Since x_S = g * (x_E - v*t_E) = 0,
t_E = x_E / v, or x_E = v * t_E. Since t_S = (t_E - v * x_E / c^2) * g,
t_S = (t_E - v^2 * t_E / c^2) * g = t_E / g.

Since t_S t_E, x_E - v * t_S 0 -- this may be what Paul meant.
Unfortunately, the expression x_E - v * t_S isn't all that meaningful
an expression, as the units are not consistent.
[*] with this expression, we are eliding the issue of when the light of
the spacecraft reaching the star actually gets to Earth, and
assuming that Earth -- somehow -- sees things instantaneously.

--
#191,
Error 16: Not enough space on file system to delete file(s)

--
Posted via a free Usenet account from http://www.teranews.com

On Thu, 30 Aug 2007 10:31:32 +0100, "George Dishman"
wrote:


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

George, as far as I'm concerned, everything you say is riddled with
mistakes.
For a start, I certainly DON'T accept that the radius varies by 12%

Tough, that is the value directly measured by ESO.

it is wrong.

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

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

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

George, presumably the interference is caused by the angle subtended by the
star.
If it is rotating, the lght from both sides will be phase shifted as they
arrive due to c+/-v.

Interferometry doesn't work in BaTh.

Probably, but interferometry works in the real
world so that is only a problem for the theory.





George, does our sun noticeably vary in size?

No, and if ESO used the interferometric technique
it would show as constant.

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

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

Interferomery will give a distorted answer.

Nope, there is no distortion introduced
by ballistic theory.

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


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

George, the technique is highly suspect at best.

Garbage, it is no more suspect than a grating.

Add variable light speed and
it becomes almost useless.

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

the star's rotation stuffs up the whole process.



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

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

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

Intensity Henry.

Same thing.

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



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

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

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

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

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

you are mssing the willusions...

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

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

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

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

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

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



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

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

That's correct

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

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

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

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

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


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

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

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


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

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

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

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

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

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

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


Oh I get it Henry, better than you. The vast
majority of the luminosity change is already
explained by radius and temperature changes so
until you remove those, any contribution from
ADoppler is unknown.

No George, you are still living in that imaginary universe in which
willusions
don't exist.

No Henry I live in the real universe, "willusions"
are your imaginary effect, but regardless your
ballistic equations not not result in "willusions"
on temperature measurements and probably not on
the radius measurement to any great extent.

George, it took astronomers many centuries to realise that they were seeing
cosmic events long after they occurred. Some of us more enlightened humans now
appreciate that everything we see is also willusory because of light's variable
speed across space.
You would still be classed as one of the primatives.


Nope, the interferometer is only concerned about
the phase across the instrument of the light that
is arriving at a particular time. In fact the
interferometer will work with single photons (like
the gratings we discussed which are really a
particular type of interferometer) and obviously
each photon only has a single speed. Ballistic
theory doesn't suggest any form of distortion for
the instrument.

All photons arriving at different speeds will adjust to the same c/n on
entering Earth's atmosphere. Their absolute wavelengths will adjust
accordingly.
I gather that inferferometry effectively detects the angle subtended by
the
star.

Yes.

Small differences in emission times and relative velocities from each
side could markedly affect the results.

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

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

You might want to consider the overall setup:

http://tinyurl.com/3dybf3

No, it wont work..



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

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

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

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

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

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

They just fiddle the equations till they get the right answer.

Nope, in real science the equations have been
published and reviewed and can't be changed.
They are the same equations used for all other
branches of acoustics.

I know what happens George, I've been there...that's how I know how much
fiddling goes on....

After all, that's just what Planck did to get his black body curve.

Nope, that's what Wien and those before him
did, they are called "empirical" laws. Planck
derived his equation from the postulate that
energy was emitted in packets each with an
energy proportional to its frequency and from
the statistics for occupancy of different modes
so he had no scope to adjust the equation at
all.

that's right....but he found the equation empirically before he applied the
statistical theory.

The derivation is published so you can
repeat his process and you will get the same
answer. The fact that the equation fits the
observations is what gives confidence that
the postulate was valid.

Sorry Henry, interferometers here on Earth get
light moving at c/n through the atmosphere
even in ballistic theory so the result is
identical with BaTh.

No George. Light travels a long way before it reaches Earth..and the
phasing
between light emitted at slightly different times will vary accordingly
when
mixed with the interfrometer signal.

The phase is relevant to each single photon Henry,
you cannot have interference between light emitted
at slightly different times because it is thermal
(black body) so uncorrelated. The interferometer
is sensitive to arrival time differences for each
photon individually, the resulting curves are the
statistical sum of the photon flux.

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

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

It ignores the different c+v from both sides.


The temperature curve is that of a relaxation
oscillator - essentially a sawtooth - with a
delay for the time for the light to reach the
surface. Acoustic theory says the motion will
be a driven and damped resonance with harmonics.
The resulting phase is not trivial to work out.

It certainly is not trivial.

Exactly, but when it is done, the models do match
the observed curves.

all willusory.....

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 Thu, 30 Aug 2007 09:47:27 +0100, "George Dishman"
wrote:


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

That distance change is negligible.

It is the radius curve and is directly
measurable for L Car.


So who is going to prove the results wrong, eh?..even if they are way out.

Results are published as they are measured
Henry, you are completely clueless about the
process. If you bothered to read the pages I
cite you would find that they measured the
angular variation and then did a best fit to
the integrated Doppler treating the distance
to the star as a free parameter. The value
they get for that distance from the fit is
close to that from other methods but not quite
the same, but it is published regardless.

What you are missing is first that their
results will be scrutinised by competing teams
who will look to find fault, and second that
their method may give a better distance measure
and be able to correct everyone else's which
means they get the credit for a major advance.

different teams of believer in constant light speed....they'r all in the same
boat really. ..and all equally confused...

You are so used to cheating your own results
by putting in "K" factors or "forces unknown"
or "speed equalisation" bodges that you have
never understood how real science is done.

Irrespective of all that, all we are concerned
with at the moment is the variation of angular
diameter because that directly affects the
luminosity and until you take it into account
your fitting process is simply wrong. For K band
the brightness change is about a fifth of the
luminosity change due to the radius variation.

George, yo can remain ignorant as long as you like...It wont affect my amazing
iscoveries.

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 Thu, 30 Aug 2007 17:17:24 +0100, "George Dishman"
wrote:


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

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

Sure, both factors operate.

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

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

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

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

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

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


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

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

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

Not if its temperature is continually changing.



Nope, the phasing is as expected, you just aren't
considering several important aspects.

I can imagine the phasing being very different in different types of
stars.
Certainly the phasing of the overtone varies considerably.

Of course, but the way it varies is correctly
predicted by the models.




Is that all you can say...'whatever' when we're discussing the basis of my
'K
factor' theory..?

I've told you before, analogies are only useful
as an aid to understanding. Until you provide
your equation, there is nothing to be understood.
Do the science first and look for analogies later.

the science should be pretty obvious to a chief engineer.
How much does a rubber ball contract when it sinks?


Nor can you have photons with negative lengths...


That's your 'wave equation' of course....
Photons are particles.


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

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

I know that George.
I'm talking about the rate of fall, which will depend on viscosity (temperature
dependent) and size (pressure dependent)


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

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

they are big assumptions...particularly as you claim there is NO temperature
equilibrium.


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

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

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

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

There are still a few areas where the models
aren't complete, AFAIK mainly in transverse
modes (acoustic waves going over the surface
of the star rather than radially) but that
is cutting edge stuff and my knowledge is
superficial.

The models are all based on willusory Einstiniana stuff and are wrong...

Sorry Henry, they match observation so they are
right by the only measure that counts in science.

they match one willusion against another.

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.

HW@....(Henri Wilson) wrote in
:

On Thu, 30 Aug 2007 12:31:39 +0000 (UTC), bz
wrote:

HW@....(Henri Wilson) wrote in
m:

not solid rubber ones.....that's what I'm talking about.

Whatever.

Is that all you can say...'whatever' when we're discussing the basis
of my 'K factor' theory..?


Henri, your 'K factor' theory died some time back when I pointed out
that a 'K factor' compression of photons implies observable effects that
are not observed.

......but they are
That is why brightness and velocity curves are usually similar in shape
but different in magnituge change.

Any effect on photons causing them to compress when crowded together
would show up as shifts in wavelength and frequency of the emission from
high intensity sources, such as lasers.

...but photon density isn't actually what CAUSES them to compress. It
just happens to occur concurrently.Photons compress if their source is
accelerating or if they change speed during travel...but they're kind of
'damped' so the movement doesn't go on forever.

Photons 'change speed' during travel when they go from one medium to
another. Yet we see no sign of photon "compression" taking place, even
when the speed change is drastic, the groups of photons and the photons
themselves change 'length' by exactly the same percentage, as George has
repeatedly pointed out. The widths of pulses from pulsars, the spacing
between the pulses and the frequencies of the pulses change by the exactly
the same amounts, likewise laser pulses traveling along optical fibers


Also, 'as the pressure goes down, the photons would decompress' just
like the rubber ball springs back when removed from the depths. Surely
the weak streams of photons we receive from those distant stars have
insufficient 'pressure' to keep the photons compressed.

You can't propose a 'non elastic compression', where the photons stay
compressed because they are already 'highly compressed' at the time of
emission by the star.

the K factor is small. maybe 10^-4

No matter what it is, it must be observed and if it is truly a 'K'
'factor', it must be constant. Perhaps it is a 'D' factor and proportional
to the density of the waether.

Also lasers can operate at very low emission rates (in fact, there are
single photon lasers) and any such effect would show up as drastic
shifts in the emission band as the laser's output power was varied.

Give up on your 'K'. It is disproved daily by millions of laser diodes
used for gigabyte fiber optical data transmission.

no Bob you have it all wrong...

A 'K' won't work because it doesn't work.


If the photons 'bunched up' the way you propose, it would cause very
strong phase shifts and keying transients, making it impossible to push
data down those fibers at the rates data is sent, every day.

If you ever have heard a radio-telegraph transmitters that has chirp
(frequency shift during turn-on) and clicks (wide keying sidebands due
to too sharp turn-on/turn-off), you will know that such a transmitter
can cause interference with communications across a wide portion of the
radio spectrum. Any attempt to transmit data at a high data rate, with
such a transmitter, would fail.

That is exactly why your 'K' factor 'photon compression' idea is dead.

You have entirely the wrong impression...

Then tell me how your 'rubber cars' look now.




--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

"The Ghost In The Machine" wrote in message
...
: In sci.physics.relativity, Androcles
:
: wrote
: on Sun, 02 Sep 2007 11:10:04 GMT
: :
:
: "The Ghost In The Machine" wrote in
message
: ...
: : In sci.physics.relativity, Androcles
: :
: : wrote
: : on Sat, 01 Sep 2007 14:03:45 GMT
: : :
: :
: : "Paul B. Andersen" wrote in message
: : ...
: : : Androcles wrote:
: : : "Paul B. Andersen" wrote in
message
: : : ...
: : : : Androcles wrote:
: : : : "Paul B. Andersen" wrote in
: : message
: : : : ...
: : : : : Henri Wilson wrote:
: : : : : Explain the 90 deg phase lag then George.
: : : : :
: : : : : Not hard at all to explain why the curves are different.
: : : : : If the radius of the star didn't change, it is obvious
: : : : : from Planck's blackbody equation that the luminosity
: : : : : variation due to the changing temperature is much
: : : : : bigger in visible light (V-band) than it is in IR (K-band
: 2.2u).
: : : :
: : : : Everything is "obvious". Obviously you are a lunatic.
: : : :
: : : : I note with a yawn that Androcles doesn't find the obvious
: : : : consequence of Planck's black body radiation law to be obvious.
: : : :
: : : : Nothing is obvious in a haze, is it?
: : :
: : : ASSistant Professor "Paul B. Andersen" of
:
: : : Agder University College (HiA)
: : : Serviceboks 422, N-4604 Kristiansand, NORWAY Tel (+47) 38 14 10 00
: Fax
: : : (+47) 38 14 10 01
: : : has executed the biggest fumble ever seen in the history of
: : : sci.physics.relativity
: : : in message
: : : ...
: : :
: : : "The spectral class [of stars] is determined by the relative
: positions
: : : and intensities
: : : of the absorption lines, and these are unaffected by a Doppler
: shift."
: : :
: : : The all time classic:
: : :
: : : "That is, we can reverse the directions of the frames
: : : which is the same as interchanging the frames,
: : : which - as I have told you a LOT of times,
: : : OBVIOUSLY will lead to the transform:
: : : t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
: : : x = (xi - v*tau)/sqrt(1-v^2/c^2)
: : : or:
: : : tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
: : : xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
: : :
: : :
: : The faster you go the longer it takes to get there, OBVIOUSLY.
: : yawn
: :
: : Here's how I look at it.
:
: [snip wrong argument]
:
: ... reach the star, d-vt = 0.
:
: The Andersen Transforms are d+vt 0.
: The faster you go the longer it takes to get there, OBVIOUSLY.
: yawn
:
: Since I was assuming a different transform:
:
: xi = (x - vt) * g
: tau = (t - vx/c^2) * g
:
: this doesn't quite work.

: However, I'd have to dig through
: the posts to see what Paul was claiming and your objections
: thereto,

That's an old worn-out transform that doesn't quite work.
The Andersen Transforms date back to 1999 to 2000 and
are the symmetric form of the Cuckoo... err... sorry... Lorentz
Transforms. Once you realise that the station comes to the train
(i.e. the Principle of Relativity) and that the K-frame that was
the "stationary" frame becomes the "moving" frame as seen by
an observer on the train (we always have observers in relativity,
never passengers) then

"That is, we can reverse the directions of the frames
which is the same as interchanging the frames,
which - as I have told you a LOT of times,
OBVIOUSLY will lead to the transform:
t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
x = (xi - v*tau)/sqrt(1-v^2/c^2)"

which is the cuckoo malformation as seen by the train observer
obviously to do. (I wrote it in the original Norwegian by adding
"to do" at the end and included "obviously".)
Obviously all that has happened is the names 't' and 'tau', 'x' and 'xi'
have been exchanged, quite, to do.

Now comes the miraculous part, to do.

"or:
tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen

Obviously, quite, to do.



: However, I'd have to dig through
: the posts to see what Paul was claiming and your objections
: thereto, and I'd highly prefer a different nomenclature anyway,
: something along the following lines.

: If E is a symbol for the coordinate frame for Earth and S
: the symbol for the coordinate frame for the spacecraft,


Aha, we are off into spacetime, how exciting! Choo-choo
trains are definitely old hat these days.



: and the spacecraft is traveling at a uniform velocity
: v away from Earth in a straight line, and with
: coincident origins, one can establish the usual Lorentz transform
:
: x_S = (x_E - v * t_E) / sqrt(1-v^2/c^2)
: t_S = (t_E - v * x_E / c^2) / sqrt(1-v^2/c^2)
:
: This transform is invertable, with a little work,
: resulting in:
:
: x_E = (x_S + v * t_S) / sqrt(1-v^2/c^2)
: t_E = (t_S + v * x_S / c^2) / sqrt(1-v^2/c^2)

There ya go, the Andersen Transforms. Well done!

Now, keeping the origo (that's Norwegian for origin and not an
oreo cookie or cuckoo with cream) of the E-frame and the S-frame
coincident as you say above, v = 0.

So with a little more work,
x_E = (x_S + 0 * t_S) / sqrt(1-0^2/c^2)
t_E = (t_S + 0 * x_S / c^2) / sqrt(1-0^2/c^2)

x_E =x_S
t_E = t_S.

:
: I can also express these using the notation
:
: (x_E,t_E)_E = ( (x_E - vt_E)*g, (t_E - v*x_E/c^2)*g)_S
: (x_S,t_S)_S = ( (x_S + vt_S)*g, (t_S + v*x_S/c^2)*g)_E
:
: where _S and _E establish which frame I'm using for the
: given coordinate point, and g = 1/sqrt(1-v^2/c^2) the
: usual gamma. I'm not sure if this is a proper tensor,
: but it does establish a mapping from E-frame to S-frame
: (and vice versa; the transform is bijective for any v in
: the open interval (-1, +1)),

As long as the spacecraft is traveling at a uniform velocity
zero away from Earth in a straight line with coincident origins.

: and I for one think

No, surely not! Can you provide any evidence to support that
absurb claim?



it helps
: clarify things, as one doesn't have to ask such questions
: as "which frame is this relevant in?".

Of course. Keeping the origins coincident really does
clarify things.



: Now one can set up various problems: the most obvious
: one is when does SR predict that the spacecraft reaches
: the star?


It doesn't, the origins coincide. You'll need to bring the
star to the coincident origins.




: More specifically, what values of x_E, t_E,
: x_S, and t_S are such that
: (x_E, t_E)_E = (d,t_E)_E [Earth sees it reaching the star][*]
: and
: (x_S, t_S)_S = (0,t_S)_S ? [The spacecraft sees itself reaching the
star]
:
: The first equality should be obvious, but the second one might not be.
: The general issue is that an astronaut cannot leave his ship, and since
: the ship defines the ship-frame, x_S = 0 for most of the interesting
: measurements, unless one postulates the astronaut holding a rod of
: a certain length (which is a slightly different problem).
:
: So now we know x_E = d, x_S = 0. Since x_S = g * (x_E - v*t_E) = 0,
: t_E = x_E / v, or x_E = v * t_E. Since t_S = (t_E - v * x_E / c^2) * g,
: t_S = (t_E - v^2 * t_E / c^2) * g = t_E / g.
:
: Since t_S t_E, x_E - v * t_S 0 -- this may be what Paul meant.
: Unfortunately, the expression x_E - v * t_S isn't all that meaningful
: an expression, as the units are not consistent.
:
:[*] with this expression, we are eliding the issue of when the light of
: the spacecraft reaching the star actually gets to Earth, and
: assuming that Earth -- somehow -- sees things instantaneously.
:
Did you want further comment on coincident origins, the value of v
or your ability to think?

The Ghost In The Machine wrote:
In sci.physics.relativity, Androcles

wrote
on Sun, 02 Sep 2007 11:10:04 GMT
:
"The Ghost In The Machine" wrote in message
...
: In sci.physics.relativity, Androcles
:
: wrote
: on Sat, 01 Sep 2007 14:03:45 GMT
: :
:
: "Paul B. Andersen" wrote in message
: ...
: : Androcles wrote:
: : "Paul B. Andersen" wrote in message
: : ...
: : : Androcles wrote:
: : : "Paul B. Andersen" wrote in
: message
: : : ...
: : : : Henri Wilson wrote:
: : : : Explain the 90 deg phase lag then George.
: : : :
: : : : Not hard at all to explain why the curves are different.
: : : : If the radius of the star didn't change, it is obvious
: : : : from Planck's blackbody equation that the luminosity
: : : : variation due to the changing temperature is much
: : : : bigger in visible light (V-band) than it is in IR (K-band
2.2u).
: : :
: : : Everything is "obvious". Obviously you are a lunatic.
: : :
: : : I note with a yawn that Androcles doesn't find the obvious
: : : consequence of Planck's black body radiation law to be obvious.
: : :
: : : Nothing is obvious in a haze, is it?
: :
: : ASSistant Professor "Paul B. Andersen" of :
: : Agder University College (HiA)
: : Serviceboks 422, N-4604 Kristiansand, NORWAY Tel (+47) 38 14 10 00
Fax
: : (+47) 38 14 10 01
: : has executed the biggest fumble ever seen in the history of
: : sci.physics.relativity
: : in message
: : ...
: :
: : "The spectral class [of stars] is determined by the relative
positions
: : and intensities
: : of the absorption lines, and these are unaffected by a Doppler
shift."
: :
: : The all time classic:
: :
: : "That is, we can reverse the directions of the frames
: : which is the same as interchanging the frames,
: : which - as I have told you a LOT of times,
: : OBVIOUSLY will lead to the transform:
: : t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
: : x = (xi - v*tau)/sqrt(1-v^2/c^2)
: : or:
: : tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
: : xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
: :
: :
: The faster you go the longer it takes to get there, OBVIOUSLY.
: yawn
:
: Here's how I look at it.

[snip wrong argument]

... reach the star, d-vt = 0.

The Andersen Transforms are d+vt 0.
The faster you go the longer it takes to get there, OBVIOUSLY.
yawn

Since I was assuming a different transform:

xi = (x - vt) * g
tau = (t - vx/c^2) * g

this doesn't quite work. However, I'd have to dig through
the posts to see what Paul was claiming and your objections
thereto, and I'd highly prefer a different nomenclature anyway,
something along the following lines.

What I claim should be quite clear from Androcles'
correct quotation, even if it is quoted out of context.

The LT as written i Einstein's 1905 paper is:
tau = (t - xv/c^2)/sqrt(1-v^2/c^2)
xi = (x - vt)/sqrt(1-v^2/c^2)
or:
t = (tau + xi*v/c^2)/sqrt(1-v^2/c^2)
x = (xi + v*tau)/sqrt(1-v^2/c^2)

Here the origin of the "Greek frame" is moving along
the positive x-axis of the "Latin frame" with speed v.

But alternatively we could let the origin of the "Latin frame"
move along the positive xi-axis of the "Greek frame".

"That is, we can reverse the directions of the frames
which is the same as interchanging the frames,
which - as I have told you a LOT of times,
OBVIOUSLY will lead to the transform:
t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
x = (xi - v*tau)/sqrt(1-v^2/c^2)
or:
tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
xi = (x + vt)/sqrt(1-v^2/c^2)"

This is but an obvious triviality, and Androcles'
'objectons thereto' are so nonsensical that they
are not worth wasting your time at.

Or mine.

Paul

"Paul B. Andersen" wrote in message
...
: The Ghost In The Machine wrote:
: In sci.physics.relativity, Androcles
:
: wrote
: on Sun, 02 Sep 2007 11:10:04 GMT
: :
: "The Ghost In The Machine" wrote in
message
: ...
: : In sci.physics.relativity, Androcles
: :
: : wrote
: : on Sat, 01 Sep 2007 14:03:45 GMT
: : :
: :
: : "Paul B. Andersen" wrote in
message
: : ...
: : : Androcles wrote:
: : : "Paul B. Andersen" wrote in
message
: : : ...
: : : : Androcles wrote:
: : : : "Paul B. Andersen" wrote
in
: : message
: : : : ...
: : : : : Henri Wilson wrote:
: : : : : Explain the 90 deg phase lag then George.
: : : : :
: : : : : Not hard at all to explain why the curves are different.
: : : : : If the radius of the star didn't change, it is obvious
: : : : : from Planck's blackbody equation that the luminosity
: : : : : variation due to the changing temperature is much
: : : : : bigger in visible light (V-band) than it is in IR (K-band
: 2.2u).
: : : :
: : : : Everything is "obvious". Obviously you are a lunatic.
: : : :
: : : : I note with a yawn that Androcles doesn't find the obvious
: : : : consequence of Planck's black body radiation law to be
obvious.
: : : :
: : : : Nothing is obvious in a haze, is it?
: : :
: : : ASSistant Professor "Paul B. Andersen"
of :
: : : Agder University College (HiA)
: : : Serviceboks 422, N-4604 Kristiansand, NORWAY Tel (+47) 38 14 10
00
: Fax
: : : (+47) 38 14 10 01
: : : has executed the biggest fumble ever seen in the history of
: : : sci.physics.relativity
: : : in message
: : : ...
: : :
: : : "The spectral class [of stars] is determined by the relative
: positions
: : : and intensities
: : : of the absorption lines, and these are unaffected by a Doppler
: shift."
: : :
: : : The all time classic:
: : :
: : : "That is, we can reverse the directions of the frames
: : : which is the same as interchanging the frames,
: : : which - as I have told you a LOT of times,
: : : OBVIOUSLY will lead to the transform:
: : : t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
: : : x = (xi - v*tau)/sqrt(1-v^2/c^2)
: : : or:
: : : tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
: : : xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
: : :
: : :
: : The faster you go the longer it takes to get there, OBVIOUSLY.
: : yawn
: :
: : Here's how I look at it.
:
: [snip wrong argument]
:
: ... reach the star, d-vt = 0.
:
: The Andersen Transforms are d+vt 0.
: The faster you go the longer it takes to get there, OBVIOUSLY.
: yawn
:
: Since I was assuming a different transform:
:
: xi = (x - vt) * g
: tau = (t - vx/c^2) * g
:
: this doesn't quite work. However, I'd have to dig through
: the posts to see what Paul was claiming and your objections
: thereto, and I'd highly prefer a different nomenclature anyway,
: something along the following lines.
:
: What I claim should be quite clear from Androcles'
: correct quotation, even if it is quoted out of context.
:
: The LT as written i Einstein's 1905 paper is:
: tau = (t - xv/c^2)/sqrt(1-v^2/c^2)
: xi = (x - vt)/sqrt(1-v^2/c^2)
: or:
: t = (tau + xi*v/c^2)/sqrt(1-v^2/c^2)
: x = (xi + v*tau)/sqrt(1-v^2/c^2)
:
: Here the origin of the "Greek frame" is moving along
: the positive x-axis of the "Latin frame" with speed v.
:
: But alternatively we could let the origin of the "Latin frame"
: move along the positive xi-axis of the "Greek frame".
:
: "That is, we can reverse the directions of the frames
: which is the same as interchanging the frames,
: which - as I have told you a LOT of times,
: OBVIOUSLY will lead to the transform:
: t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
: x = (xi - v*tau)/sqrt(1-v^2/c^2)
: or:
: tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
: xi = (x + vt)/sqrt(1-v^2/c^2)"
:
: This is but an obvious triviality, and Androcles'
: 'objectons thereto' are so nonsensical that they
: are not worth wasting your time at.
:
: Or mine.
:
: Paul

S-frame:
Let c = 1, v = 0.5, x = 0.866.
t = x/v = 1.732 years

E-frame:
tau = 1.655 years t




S-frame:
Let c = 1, v = 0.866, t = 1 year.
x = vt = 0.866 as before

E-frame:
tau = (1.75 /1^2)/ 0.5 = 3.5 years t

The faster you go the longer it takes to get there.
Check with a spreadsheet:
x v t tau t/tau
0.866 0.100 8.660 1.092 7.930
0.866 0.200 4.330 1.197 3.616
0.866 0.300 2.887 1.321 2.186
0.866 0.400 2.165 1.469 1.474
0.866 0.500 1.732 1.655 1.047
0.866 0.600 1.443 1.900 0.760
0.866 0.700 1.237 2.249 0.550
0.866 0.800 1.083 2.821 0.384
0.866 0.866 1.000 3.500 0.286
0.866 0.900 0.962 4.082 0.236
0.866 0.990 0.875 13.167 0.066
0.866 0.999 0.867 41.717 0.021


This is but an obvious stupidity, and Tusseladd's
'idiocies thereto' are so nonsensical that they
are not worth wasting your time at -- but what else is
there to do except laugh at stupidity and incompetence?