Why are the 'Fixed Stars' so FIXED?

"Dr. Henri Wilson" HW@.... wrote in message
...
On Wed, 26 Sep 2007 05:32:46 -0700, George Dishman
wrote:
On 26 Sep, 11:49, HW@....(Dr. Henri Wilson) wrote:
On Wed, 26 Sep 2007 01:11:39 -0700, George Dishman
wrote:
On 25 Sep, 22:53, HW@....(Dr. Henri Wilson) wrote:
....
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.

That's a funny paragraph.
'opacity' ...'specific frequency' ...I didn't know these had the same
physical
dimensions...

Why would you imagine they would? The opacity
of the gas just above the photosphere is higher
at the specific frequency we call hydrogen Alpha
than at other frequencies between such resonances.
You even claimed you had taken pictures in H alpha.

Sorry I misread your statement...
It's OK..

I assume you read "higher than" instead
of "higher at that". No problem.

...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.

If the shift is only 0.01%, so what?

So the amount of energy that falls outside a
bandpass filter from 2000 to 2400 nm because
of the shift is negligible hence the temperature
measurement based on that value is valid.

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.

Many stars do not exhibit black body curves so how would you now if they
were
shifted or not.

The shift is measured from specific lines,
you cannot measure a shift of 0.01% in the
continuum which is never exactly a black
body because of the finite thickness. This
is all obvious stuff Henry.

George, let's get this straight.

I think that would be a good idea, but we
have been over it several times and you
continue to ignore the numbers. Until you
start listening to my points and responding
to them, nothing will be resolved.

You say temperature is estimated by comparing the energy arriving in the
two
fliter bands and assuming a black body curve.

Almost, it doesn't assume a perfect black body
but the typical curve which is the integral
over the depth. The measurements are calibrated
for the complex shape of the filters and that
in practice absorb this effect.

I say the average light in the two bands can come from slightly different
layers with different radial velocities.

That doesn't work when you put the numbers in
though. The worst case would be if one band
was from the top of the photosphere and the
other was from the bottom. At most they are
of the order of 400km apart. The photosphere
moves by millions of km so the difference in
speed between the top and bottom is tiny, nil
if the width of the layer stays constant. If
it changed from 350km to 450km between the
times of maximum and minimum radius, that's
only a speed difference of 100km in 35 days
compared to something on the order of 30km/s
mean speed. In reality, the light comes from
almost the same depth in both bands.

ADoppler can shift the light by much
more than 0.01%

What it _might_ be is irrelevant, we know the
_actual_ shift and it is around 0.01%, my
example typical figure of 30km/s divided by c
regardless of what causes it. You can look up
the actual figure for L Car if you like.

and might easily lead to false ratios in the two bands.

As a handwaving exercise, that would appear
right but first you need to do the physics. The
degree of error is given by noting the shift is
0.01% of the frequency of the band edge and as
I showed before that is at most a fraction of a
degree. The temperature is observed to change
typically by more than 1000 degrees so an error
of less than 1 degree is completely negligible
and the temperature swing measured is valid.

I have explained all this to you several times
now so don't waste my time with more handwaving,
address the actual numbers if you still want to
argue.

George

On Fri, 28 Sep 2007 16:37:09 +0100, "George Dishman"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
On Wed, 26 Sep 2007 05:32:46 -0700, George Dishman

shifted or not.

The shift is measured from specific lines,
you cannot measure a shift of 0.01% in the
continuum which is never exactly a black
body because of the finite thickness. This
is all obvious stuff Henry.

George, let's get this straight.

I think that would be a good idea, but we
have been over it several times and you
continue to ignore the numbers. Until you
start listening to my points and responding
to them, nothing will be resolved.

You say temperature is estimated by comparing the energy arriving in the
two
fliter bands and assuming a black body curve.

Almost, it doesn't assume a perfect black body
but the typical curve which is the integral
over the depth. The measurements are calibrated
for the complex shape of the filters and that
in practice absorb this effect.

I say the average light in the two bands can come from slightly different
layers with different radial velocities.

That doesn't work when you put the numbers in
though. The worst case would be if one band
was from the top of the photosphere and the
other was from the bottom. At most they are
of the order of 400km apart. The photosphere
moves by millions of km so the difference in
speed between the top and bottom is tiny, nil
if the width of the layer stays constant. If
it changed from 350km to 450km between the
times of maximum and minimum radius, that's
only a speed difference of 100km in 35 days
compared to something on the order of 30km/s
mean speed. In reality, the light comes from
almost the same depth in both bands.

there's no point in plucking figures out of the air just to make your theory
look good....

ADoppler can shift the light by much
more than 0.01%

What it _might_ be is irrelevant, we know the
_actual_ shift and it is around 0.01%, my
example typical figure of 30km/s divided by c
regardless of what causes it. You can look up
the actual figure for L Car if you like.

Forget L Car....what about all the other stars?

and might easily lead to false ratios in the two bands.

As a handwaving exercise, that would appear
right but first you need to do the physics. The
degree of error is given by noting the shift is
0.01% of the frequency of the band edge and as
I showed before that is at most a fraction of a
degree. The temperature is observed to change
typically by more than 1000 degrees so an error
of less than 1 degree is completely negligible
and the temperature swing measured is valid.

I have explained all this to you several times
now so don't waste my time with more handwaving,
address the actual numbers if you still want to
argue.

You can now read about how Sagnac fully supports the BaTh. See latest thread.

George



Henri Wilson. ASTC,BSc,DSc(T)

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

"Dr. Henri Wilson" HW@.... wrote in message
...
On Fri, 28 Sep 2007 16:37:09 +0100, "George Dishman"
wrote:
"Dr. Henri Wilson" HW@.... wrote in message
. ..
On Wed, 26 Sep 2007 05:32:46 -0700, George Dishman

The shift is measured from specific lines,
you cannot measure a shift of 0.01% in the
continuum which is never exactly a black
body because of the finite thickness. This
is all obvious stuff Henry.

George, let's get this straight.

I think that would be a good idea, but we
have been over it several times and you
continue to ignore the numbers. Until you
start listening to my points and responding
to them, nothing will be resolved.

You say temperature is estimated by comparing the energy arriving in the
two fliter bands and assuming a black body curve.

Almost, it doesn't assume a perfect black body
but the typical curve which is the integral
over the depth. The measurements are calibrated
for the complex shape of the filters and that
in practice absorb this effect.

I say the average light in the two bands can come from slightly
different
layers with different radial velocities.

That doesn't work when you put the numbers in
though. The worst case would be if one band
was from the top of the photosphere and the
other was from the bottom. At most they are
of the order of 400km apart. The photosphere
moves by millions of km so the difference in
speed between the top and bottom is tiny, nil
if the width of the layer stays constant. If
it changed from 350km to 450km between the
times of maximum and minimum radius, that's
only a speed difference of 100km in 35 days
compared to something on the order of 30km/s
mean speed. In reality, the light comes from
almost the same depth in both bands.

there's no point in plucking figures out of the air just to make your
theory
look good....

I gave you the accurate figures and the paper from
which they were taken when we discussed it before,
I just couldn't be bothered to dig them out again
when you pay no attention to real physics anyway.

ADoppler can shift the light by much
more than 0.01%

What it _might_ be is irrelevant, we know the
_actual_ shift and it is around 0.01%, my
example typical figure of 30km/s divided by c
regardless of what causes it. You can look up
the actual figure for L Car if you like.

Forget L Car....what about all the other stars?

L Car is typical.

and might easily lead to false ratios in the two bands.

As a handwaving exercise, that would appear
right but first you need to do the physics. The
degree of error is given by noting the shift is
0.01% of the frequency of the band edge and as
I showed before that is at most a fraction of a
degree. The temperature is observed to change
typically by more than 1000 degrees so an error
of less than 1 degree is completely negligible
and the temperature swing measured is valid.

I have explained all this to you several times
now so don't waste my time with more handwaving,
address the actual numbers if you still want to
argue.

You can now read about how Sagnac fully supports the BaTh.

So you've found another way to get it wrong have you.
We have analysed it already Henry using your own
diagram.

See latest thread.

I can't see any new threads from you in sci.astro,
and if it only relates to Sagnac then it's probably
not appropriate here anyway. We were discussing
Cepheids which is on-topic. I'll have a look on
Google instead.

George

"George Dishman" wrote in message
oups.com...
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@4 ax.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?
....
Your numbers are completely wrong...

Haven't you been able to do this yet Henry? I'm
still waiting.

George

On Fri, 28 Sep 2007 01:12:05 -0700, George Dishman
wrote:

On 27 Sep, 22:55, HW@....(Dr. Henri Wilson) wrote:
On Thu, 27 Sep 2007 01:16:58 -0700, George Dishman wrote:

explained by QED.

Oh crap.
Of course light exists is some form during its lifetime of travel.
That form can be described in terms of OUR physical 3D, 1T universe.
This is pure physics, not philosophy.

No, physics is about what can be measured so
by definition, what happens when it isn't
measured is not physics. What we can say is
that certain properties are conserved which
implies that they have the same values when
not being measured, but that is only an
implication, it cannot be confirmed obviously.

George, you've been in engineering too long....

The Einstein claim that all starlight in the universe travels miraculously at
exactly c wrt litlte planet Earth is philosophy....the fairyland type...

Of course, your pathetic strawman is pure
fairyland stuff. In the real world, SR says
the speed is c in _all_inertial_ frames, and
Earth isn't inertial, while there are an
infinite number of inertial frames.

SR is bull.
Sagnac proves that.

And I have explained that the phase difference
through the two light paths for a single photon
can only depend on the speed over the last few
wavelengths prior to the surface of the telescopes.
Any speed variations prior to that are common to
both paths since we are talking about just one
photon. There is no point in repeating what you
said when I have already told you why it does not
have any effect.

...and I have tried to explain to you that one photon cannot reveal anything
about any radius change.

And I have told you that it is the contrast ratio of
the overlapping interference patterns that is used
to determine the radius. Interference applies to
each photon individually so is unaffected by the
speed in space but it is the cumulative distribution
that gives us the information.

You can believe that if you want to



George, YOUR equation is trivial. The hard part is finding n(s) and r(s).

My equation is the one given by ballistic theory.
For n(s) and r(s) you need to look at material
science as well. For example to predict the
refractive index of diamond, you need to
understand the crystal structure, and that isn't
the province of your theory.

relate to actual situations so for example for
the pulsars we looked at, the value of n(s) is
measured using the frequency dependence. In
reality, r(s) cannot be measured since it
doesn't exist.

your opinion....

And that of all scientists, but it means you
can't find the work done for you, you need to
work it out yourself.

Time is not unlimited George.

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

..and dv/dt will also be = A * (B - v)
since vc.

Yes, and it would be a good excercise for you to
see if you can work ou why I chose to use dv/ds
instead of dv/dt. You'll need to learn some
calculus to do that.



Obviously n(s) and r(s) are much easier to consider as functions of distance,
rather than time.

Yes, that's one reason and perfectly valid, but
there is another that comes purely from calculus.
It's a bit more subtle and not a rigid requirement,
it just make life a bit easier.

Well naturally if you are dealing with functions of distance you wont try to
differentiate wrt time.



George, I have a mind that does these things automatically. I don't need any
equations until I get the model right.

The equations _are_ the model Henry, no equations
means no theory.

You being an engineer require an equation BEFORE you get the model right.
Physicists provide engineers with equations.

Exactly! So if you want to pretend you are a
physicist, _you_ should be providing _me_ with
rthe equations, not the other way round as it
is at the moment.

have a peek at these george: http://www.users.bigpond.com/hewn/sagnac.jpg

hahahahahaha!

George, stop raving and derive n(s) and r(s) for the various situations please.

I have already given you the link to the dispersion
equation for n(s) from the Jodrell Bank site. There
is no equation for r(s) but if you remember we found
an upper value for the mean of r(s) from the pulsar
data. This is supposed to be your theory so you have
to write it yourself Henry, but I can tell you how I
would approach it if I were you.

There is NO clearcut equation for either.

http://www.jb.man.ac.uk/~pulsar/tuto...ut/node14.html

There are many different situations regarding the common EM sphere around
binary pairs.

I am talking about the ISM, not the region near
the star.

'Type 2' extinction ...or unification.

I suppose it wouldn't be too hard to describe what YOU are likening to a
refractive index, n(s), as something like: 1-n = (1-n(o)).K/(s^3)....but that
would be an oversimplification.

The equation is given on the page above.

r(s) would be hard to describe since it involves statistics. Light randomly
speeds up and slows down as it travels....
We can regrad the universe as being turbulent like any rare gas.
However, we can state that type(2) unification AFTER LEAVING THE STAR'S SPHERE
requires that c+v+u - c-v+u. in other words the v - zero with distance. After
unification, the velocity 'u' will also vary depending on the properties of the
particular region of space the light happens to be in.

The same applies within the sphere but you cannot
assume r(s) is constant. An equation deriving r(s)
from density could give you the overall formula by
noting that the stellar wind is likely to fall off
roughly as the inverse square (not quite since the
material is being slowed by gravity but good for
a first approximation).

There is an important point that I must enlarge upon. In the case of a long
period orbiting star, the sphere around that star moves pretty well in phase
with it. All light leaves the sphere at about c wrt the star's barycentre.
IFall the star's emitted light experiences a speed change in escaping from the
sphere (including the gravity field) that will merely show up as a component of
proper motion towards earth. The BaTh predicted brightness curve will not be
affected.
Get it?


To get r(s) finally, you need to work out the number
of such interactions per unit distance as a function
of the density in a similar fashion to the pulsar
formula I showed you for the refractive index.

This is still the 'refractive index' approach.

It is a particulate approach, nothing more.

I maintain that unification involves MORE than ordinary matter.

Typically the particles would be assumed to be
the species observed, mostly electrons, protons,
alpha particles etc. but there is no reason why
it shouldn't apply to particles other than
normal matter, all you need is the value of M
which is the coefficient of the momentum to
speed function for the particle.

I think there is a lot more to this than we know about at present.

If a photon is a discreet particle, it must interact with its surroundings in
order to change speed ...up or down...

Yep, and the above deals with any particulate
form of "surroundings".

I still like the idea that a photon is basically a rapidly rotating pair of
charges carving out a helix or some kind of spatial patern as they travel.
Polarization has to be explained though.

Polarisation is easy, your model is similar to a
conventional circularly polarised signal and two
can be combined to give linear polarisation. Where
you have a problem is conceptually since your are
trying to give a particle-based theory so there
are no fields, the photon _is_ the field, so you
end up with photons giving off photons, and
experimentally you have to explain why we cannot
detect the individual charges.

Well why don't you think positively about the theory instead of knocking it.
This is something like Len Gaasenbeek's helical wave photon idea, which you
might recall. Len seems to have left us...

There is a problem unfortunately, if you start with
Vi 0 (e.g. a "c-v" photon) then the particle moves
more slowly and the recoil means the same process
speeds up the photon as you want, however the particle
still gets a positive speed and the emitted photon has
speed c after the interaction. Solving that one is
your task, this is supposed to be your theory after all
so I can't do it all for you.

At least you are now trying rather than rejecting it outright....
Thanks George...

Oh I reject it outright of course, Sagnac proves
the whole idea is wrong from the start, but I am
happy to teach you how to do physics for yourself.

http://www.users.bigpond.com/hewn/sagnac.jpg

Hahahahohohoho!

WHO'S TEACHING WHO NOW GEORGE?

So the question is can you do the physics and solve
the problem? I have given you a hint, if you use
the same approach of breaking the equation down
logically as I did to write equation [3], you
will find the solution. The key is that you apply
the method to work out the equation for the
momentum of a photon as a function of its speed.
See if you can do it yourself this time, if not
I'll tell you the answer and show you how to do
it again.

I don't have time at present..

George

Henri Wilson. ASTC,BSc,DSc(T)

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

On Sat, 29 Sep 2007 14:36:16 +0100, "George Dishman"
wrote:


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


I have explained all this to you several times
now so don't waste my time with more handwaving,
address the actual numbers if you still want to
argue.

You can now read about how Sagnac fully supports the BaTh.

So you've found another way to get it wrong have you.
We have analysed it already Henry using your own
diagram.

hahahahohohoho!
Have another look George.
SR is DEAD.

See latest thread.

I can't see any new threads from you in sci.astro,
and if it only relates to Sagnac then it's probably
not appropriate here anyway. We were discussing
Cepheids which is on-topic. I'll have a look on
Google instead.

I didn't send it to sci.astro..

George




Henri Wilson. ASTC,BSc,DSc(T)

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

On 30 Sep, 00:16, HW@....(Clueless Henri Wilson) wrote:
On Sat, 29 Sep 2007 14:36:16 +0100, "George Dishman"
wrote:



"Dr. Henri Wilson" HW@.... wrote in message
.. .
I have explained all this to you several times
now so don't waste my time with more handwaving,
address the actual numbers if you still want to
argue.

You can now read about how Sagnac fully supports the BaTh.

So you've found another way to get it wrong have you.
We have analysed it already Henry using your own
diagram.

hahahahohohoho!
Have another look George.

Ihave, it is plagiarised from my post of the
28th November 2005, but that's OK by me, I am
happy you have confirmed my result for yourself.

SR is DEAD.

Your diagram confirms it is correct, the post
below identifies your error.

See latest thread.

http://tinyurl.com/2q523r

George

"Clueless Henri Wilson" HW@.... wrote in message
...
On Fri, 28 Sep 2007 01:12:05 -0700, George Dishman
wrote:
On 27 Sep, 22:55, HW@....(Clueless Henri Wilson) wrote:
On Thu, 27 Sep 2007 01:16:58 -0700, George Dishman
wrote:
....
And I have explained that the phase difference
through the two light paths for a single photon
can only depend on the speed over the last few
wavelengths prior to the surface of the telescopes.
Any speed variations prior to that are common to
both paths since we are talking about just one
photon. There is no point in repeating what you
said when I have already told you why it does not
have any effect.

...and I have tried to explain to you that one photon cannot reveal
anything
about any radius change.

And I have told you that it is the contrast ratio of
the overlapping interference patterns that is used
to determine the radius. Interference applies to
each photon individually so is unaffected by the
speed in space but it is the cumulative distribution
that gives us the information.

You can believe that if you want to

I have no choice, it is observed.

George, YOUR equation is trivial. The hard part is finding n(s) and
r(s).

My equation is the one given by ballistic theory.
For n(s) and r(s) you need to look at material
science as well. For example to predict the
refractive index of diamond, you need to
understand the crystal structure, and that isn't
the province of your theory.

relate to actual situations so for example for
the pulsars we looked at, the value of n(s) is
measured using the frequency dependence. In
reality, r(s) cannot be measured since it
doesn't exist.

your opinion....

And that of all scientists, but it means you
can't find the work done for you, you need to
work it out yourself.

Time is not unlimited George.

It should take less than fifteen minutes Henry,
and probably less time that it will take you to
reply to this post with yet another evation.
Your stalling is noted.

We know the light will speed up if it is too slow
and slow down if it is too fast so if v is large,
the rate of change is negative while if the speed
is low the rate is positive. That means v enters
with a negative sign. The simplest form is first
order and that will be at worst an adequate
approximation as long as v c, and may be exact,
so the basic form must be:

dv/ds = A * (B - v)

..and dv/dt will also be = A * (B - v)
since vc.

Yes, and it would be a good excercise for you to
see if you can work ou why I chose to use dv/ds
instead of dv/dt. You'll need to learn some
calculus to do that.



Obviously n(s) and r(s) are much easier to consider as functions of
distance,
rather than time.

Yes, that's one reason and perfectly valid, but
there is another that comes purely from calculus.
It's a bit more subtle and not a rigid requirement,
it just make life a bit easier.

Well naturally if you are dealing with functions of distance you wont try
to
differentiate wrt time.

No, the second reason is that it makes it easier to
solve the equation analytically using "integration
by substitution". Remember that from school Henry?

George, I have a mind that does these things automatically. I don't need
any
equations until I get the model right.

The equations _are_ the model Henry, no equations
means no theory.

You being an engineer require an equation BEFORE you get the model
right.
Physicists provide engineers with equations.

Exactly! So if you want to pretend you are a
physicist, _you_ should be providing _me_ with
rthe equations, not the other way round as it
is at the moment.

have a peek at these george: http://www.users.bigpond.com/hewn/sagnac.jpg

hahahahahaha!

Yes, that was my reaction, you got it wrong again.

George, stop raving and derive n(s) and r(s) for the various
situations please.

I have already given you the link to the dispersion
equation for n(s) from the Jodrell Bank site. There
is no equation for r(s) but if you remember we found
an upper value for the mean of r(s) from the pulsar
data. This is supposed to be your theory so you have
to write it yourself Henry, but I can tell you how I
would approach it if I were you.

There is NO clearcut equation for either.

http://www.jb.man.ac.uk/~pulsar/tuto...ut/node14.html

There are many different situations regarding the common EM sphere
around
binary pairs.

I am talking about the ISM, not the region near
the star.

'Type 2' extinction ...or unification.

"Speed unification", extinction is something
completely different.

I suppose it wouldn't be too hard to describe what YOU are likening to a
refractive index, n(s), as something like: 1-n = (1-n(o)).K/(s^3)....but
that
would be an oversimplification.

The equation is given on the page above.

r(s) would be hard to describe since it involves statistics. Light
randomly
speeds up and slows down as it travels....
We can regrad the universe as being turbulent like any rare gas.
However, we can state that type(2) unification AFTER LEAVING THE STAR'S
SPHERE
requires that c+v+u - c-v+u. in other words the v - zero with
distance. After
unification, the velocity 'u' will also vary depending on the properties
of the
particular region of space the light happens to be in.

The same applies within the sphere but you cannot
assume r(s) is constant. An equation deriving r(s)
from density could give you the overall formula by
noting that the stellar wind is likely to fall off
roughly as the inverse square (not quite since the
material is being slowed by gravity but good for
a first approximation).

There is an important point that I must enlarge upon. In the case of a
long
period orbiting star, the sphere around that star moves pretty well in
phase
with it. All light leaves the sphere at about c wrt the star's barycentre.
IFall the star's emitted light experiences a speed change in escaping from
the
sphere (including the gravity field) that will merely show up as a
component of
proper motion towards earth. The BaTh predicted brightness curve will not
be
affected.
Get it?

Yes, the concept has no effect, as I told you when
you first suggested it.

Now stop stalling and work out the formula for r(s).

To get r(s) finally, you need to work out the number
of such interactions per unit distance as a function
of the density in a similar fashion to the pulsar
formula I showed you for the refractive index.

This is still the 'refractive index' approach.

It is a particulate approach, nothing more.

I maintain that unification involves MORE than ordinary matter.

Typically the particles would be assumed to be
the species observed, mostly electrons, protons,
alpha particles etc. but there is no reason why
it shouldn't apply to particles other than
normal matter, all you need is the value of M
which is the coefficient of the momentum to
speed function for the particle.

I think there is a lot more to this than we know about at present.

I think you are incapable of writing down a simple
piece of algebra and are looking for excuses.

If a photon is a discreet particle, it must interact with its
surroundings in
order to change speed ...up or down...

Yep, and the above deals with any particulate
form of "surroundings".

I still like the idea that a photon is basically a rapidly rotating pair
of
charges carving out a helix or some kind of spatial patern as they
travel.
Polarization has to be explained though.

Polarisation is easy, your model is similar to a
conventional circularly polarised signal and two
can be combined to give linear polarisation. Where
you have a problem is conceptually since your are
trying to give a particle-based theory so there
are no fields, the photon _is_ the field, so you
end up with photons giving off photons, and
experimentally you have to explain why we cannot
detect the individual charges.

Well why don't you think positively about the theory instead of knocking
it.

Because your suggestion is nothing new, virtually a
copy of conventional physics.

There is a problem unfortunately, if you start with
Vi 0 (e.g. a "c-v" photon) then the particle moves
more slowly and the recoil means the same process
speeds up the photon as you want, however the particle
still gets a positive speed and the emitted photon has
speed c after the interaction. Solving that one is
your task, this is supposed to be your theory after all
so I can't do it all for you.

At least you are now trying rather than rejecting it outright....
Thanks George...

Oh I reject it outright of course, Sagnac proves
the whole idea is wrong from the start, but I am
happy to teach you how to do physics for yourself.

http://www.users.bigpond.com/hewn/sagnac.jpg

Hahahahohohoho!

http://www.georgedishman.f2s.com/Hen...wavelength.png

WHO'S TEACHING WHO NOW GEORGE?

I am teaching you.

So the question is can you do the physics and solve
the problem? I have given you a hint, if you use
the same approach of breaking the equation down
logically as I did to write equation [3], you
will find the solution. The key is that you apply
the method to work out the equation for the
momentum of a photon as a function of its speed.
See if you can do it yourself this time, if not
I'll tell you the answer and show you how to do
it again.

I don't have time at present..

It would take minutes, you are incapable.

George

On Mon, 01 Oct 2007 00:56:59 -0700, George Dishman
wrote:

On 30 Sep, 00:16, HW@....(Clueless Henri Wilson) wrote:
On Sat, 29 Sep 2007 14:36:16 +0100, "George Dishman"
wrote:



"Dr. Henri Wilson" HW@.... wrote in message
.. .
I have explained all this to you several times
now so don't waste my time with more handwaving,
address the actual numbers if you still want to
argue.

You can now read about how Sagnac fully supports the BaTh.

So you've found another way to get it wrong have you.
We have analysed it already Henry using your own
diagram.

hahahahohohoho!
Have another look George.

Ihave, it is plagiarised from my post of the
28th November 2005, but that's OK by me, I am
happy you have confirmed my result for yourself.

SR is DEAD.

Your diagram confirms it is correct, the post
below identifies your error.

See latest thread.

http://tinyurl.com/2q523r

There is no error George....get used to it.

....and there's more coming....



George

Henri Wilson. ASTC,BSc,DSc(T)

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

On Mon, 1 Oct 2007 21:36:15 +0100, "George Dishman"
wrote:


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

Obviously n(s) and r(s) are much easier to consider as functions of
distance,
rather than time.

Yes, that's one reason and perfectly valid, but
there is another that comes purely from calculus.
It's a bit more subtle and not a rigid requirement,
it just make life a bit easier.

Well naturally if you are dealing with functions of distance you wont try
to
differentiate wrt time.

No, the second reason is that it makes it easier to
solve the equation analytically using "integration
by substitution". Remember that from school Henry?





Exactly! So if you want to pretend you are a
physicist, _you_ should be providing _me_ with
rthe equations, not the other way round as it
is at the moment.

have a peek at these george: http://www.users.bigpond.com/hewn/sagnac.jpg

hahahahahaha!

Yes, that was my reaction, you got it wrong again.

Then have a peek at: www.users.bigpond.com/hewn/ringgyro1.jpg


There are many different situations regarding the common EM sphere
around
binary pairs.

I am talking about the ISM, not the region near
the star.

'Type 2' extinction ...or unification.

"Speed unification", extinction is something
completely different.

Not terribly different...but I prefer 'unification'

...but your equation applies more to conventional extinction.


There is an important point that I must enlarge upon. In the case of a
long
period orbiting star, the sphere around that star moves pretty well in
phase
with it. All light leaves the sphere at about c wrt the star's barycentre.
IFall the star's emitted light experiences a speed change in escaping from
the
sphere (including the gravity field) that will merely show up as a
component of
proper motion towards earth. The BaTh predicted brightness curve will not
be
affected.
Get it?

Yes, the concept has no effect, as I told you when
you first suggested it.

Now stop stalling and work out the formula for r(s).

George I have been too busy over the last couple of days proving Einstein wrong
yet again.
Sagnac Effect has nothing whatsoever to do with relativity. It is a purely
ballistic phenomenon.



I think there is a lot more to this than we know about at present.

I think you are incapable of writing down a simple
piece of algebra and are looking for excuses.

George I've just produced a few lines of algebra that make you and your
colleagues look completely stupid.

You have been completely wrong about Sagnac. ..... why should anything else
you say have any credibility?


Polarisation is easy, your model is similar to a
conventional circularly polarised signal and two
can be combined to give linear polarisation. Where
you have a problem is conceptually since your are
trying to give a particle-based theory so there
are no fields, the photon _is_ the field, so you
end up with photons giving off photons, and
experimentally you have to explain why we cannot
detect the individual charges.

Well why don't you think positively about the theory instead of knocking
it.

Because your suggestion is nothing new, virtually a
copy of conventional physics.

there isn't any conventional theory about the structure of a photon.


Oh I reject it outright of course, Sagnac proves
the whole idea is wrong from the start, but I am
happy to teach you how to do physics for yourself.

http://www.users.bigpond.com/hewn/sagnac.jpg

Hahahahohohoho!

http://www.georgedishman.f2s.com/Hen...wavelength.png

WHO'S TEACHING WHO NOW GEORGE?

I am teaching you.

See how ring gyros work according to BaTh George....

So the question is can you do the physics and solve
the problem? I have given you a hint, if you use
the same approach of breaking the equation down
logically as I did to write equation [3], you
will find the solution. The key is that you apply
the method to work out the equation for the
momentum of a photon as a function of its speed.
See if you can do it yourself this time, if not
I'll tell you the answer and show you how to do
it again.

I don't have time at present..

It would take minutes, you are incapable.

My time is valuable.
After all, I'm the person who proved Einsteinand all his followers wrong with a
simple .jpg image.

George


Henri Wilson. ASTC,BSc,DSc(T)

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