Why are the 'Fixed Stars' so FIXED?

On Thu, 22 Feb 2007 11:23:57 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Wed, 21 Feb 2007 22:34:29 -0000, "George Dishman"

wrote:
...
There you are Henry, we have derived an upper limit
on the extinction distance from the published data.
Now you understand what I was driving at, and
hopefully you also realise I really did understand
your model all along :-)

While some stars may have more or less dense plasma
around them, in general the distance should be around
that sort of level for all and shorter for stars with
a dense plasma. Note that it is much less than the
distance to the heliopause for the Sun.

George, I don't have a firm view as to why my distances are always shorter
than
the actual ones....but there must obviously be a simple explanation.

The simple explanation is that SR is correct. From
your point of view though, as light passes through
a plasma we know it is affected and that could cause
some change to the speed. The obvious explanation
would be that absorption and re-emission at each
atom encountered immediately changes the speed to c
relative to that atom, but that would eliminate any
effects so your problem is why the extinction distance
isn't the mean path length.

George, this is the picture.
We have a neutron star rotating very rapidly and at the same time orbiting a
dwarf star.
Some kind of radiation, presumeably magnetic, is emitted by the neutron star.
We are assuming its speed wrt Earth varies between about c+/- 0.00009.

My theory says that for the pulses to be observed the way they are, there must
be some kind of light speed unification taking place within one lightday of the
system barycentre. Its speed approaches c in that time. Both 1.00009 c and
0.99991c become c.
If an inverse square law is involved, most of the change must occur in much
less than 1 day.
This is perfectly in accordance with my concept of an EM FOR surrounding large
mass centres. It is not a plain 'gravity' effect. That happens separately and
shows up as Shapiro delay.

The fact that so many brightness curves are reproducable using BaTh is
enough
to keep me convinced I'm right.

I think other factors are operating here.

There are no "other factors" in Ritzian theory to
operate aside from those already in your program.
You still need to fix that bug in the velocity
curve though.

There is no bug.

See my other post for details.

Circular orbits can appear slightly elliptical and vice versa.

Perhaps, but whether the distortion caused by variable
speed exactly eliminates that caused by Kepler's Second
Law is something you should show mathematically, and I
don't believe you can do that. As a result I think you
will find there remains a slight distortion even for
your best fit.

I can enlarge the curve and superimpose a sinewave on it.
I will do that just for you.

"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."

Hmm but a genius in physics is unlikely to need to get
the dunces to integrate an exponential for him. Remember
your "challenge" that I solved in a few lines?

George you solved the wrong problem.

The integral is of hte form Total time= intgrl [1/(1+Ae^-kt)].dt (c=1)

A solution is: t +log(1+Ae^-kt) between 0 and t.

I found a simple way to closely approximate the integral using the sum of a GP
instead...it is also faster to run.

the terms are
1/(1+0.00009),1/(1+0.00009X),1/(1+0.00009X2),1/(1+0.00009X3).........1/(1+0.00009X^n)
Since the 0.00009 is small, this can be closely approximated with:
(1-0.00009),(1-0.00009X),(1-0.00009X2)...........................(1-0.00009X^n)
The sum is (n+1) -(1-0.00009)*(X^n-1)/(X-1) wherer n is the number of light
days and X is the unification rate (eg., 0.99995 per Lday)

That is really your biggest problem, you don't seem to
have the familiarity with maths that you need to follow
a lot of the arguments.

Now you're starting to sound like geesey....

At the moment you seem to be
struggling with the wavelength to velocity conversion
for your blue line for example.

I'M not....YOU are.

Anyway, see if my latest
attempt to explain it lets the penny drop and we'll see
where that takes your program.

..


George


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On Thu, 22 Feb 2007 10:17:35 -0000, "George Dishman"
wrote:


Most of our views on this are now in accord, I only
address the speed issue here and maybe pick up some
other minor points separately later.

First I'll take one paragraph from later on;


Your method doesn't take the effect of
the initial speed difference into account.

Don't be silly George, Of course it does. That's the whole basis of the
calculation.
The radial speed at each point around the orbit is c + vcos(A)

I said before you could treat cos(A) as being always 1.

A is a function of time George. I pointed that out way back.

I was thinking there of the angle between the line of
sight and the line between the barycentres. Your angle
is between to have used the line of sight and the
velocity which of course is essential but if you make
it the angle between the velocity and a line joining
the barycentres then there will be a negligible error,
essentially the view from infinity, and it will work
at zero distance to allow comparison with the conventional
model.

Thanks to Jeff Root for pointing out my misunderstanding
of your definition.

You are still completely misunderstanding the whole thing.


"Henri Wilson" HW@.... wrote in message
.. .
On Wed, 21 Feb 2007 18:35:50 -0000, "George Dishman"
wrote:
...
I did explain Henry, at the critical distance the
gap between pulses is zero so your program should
report a value of c for the observed velocity curve
but the peak is the same height as the true value
which you entered as 0.0009. That's wrong by a
factor of 11000.

I think I know what you are trying to say here George.

At the critical distance, SOME pulses arrive together not ALL of them.
that is
because a cincave section of the orbit is such tat a large group of pulses
will
arrive at a distant point over a very short time interval.
They will have started out with a range of speeds; that's why some catch
up
with the others.

Yes.

After extinction, they will all be traveling at about c wrt the source BUT
their wavelengths will have changed so that their source speeds will still
appear to be the correct ones, when measured with a grating at the
observer
distance..

No. We are not using a grating. Individual pulses have
their time of arrival noted against an atomic clock.
Remember they are 2.95 ms apart so the 'wavelength' is
885 km.

The inverse of the time between arrivals is the pulse
repetion frequency. That frequency is what is turned
into the published orbital parameters and is what give
the 339 Hz +/- 30 mHz values.

That's due to normal doppler 'bunching'. BaTh bunching is virtually the same.

So my graph shows the 'no extinction' case...because I say extinction
makes no
difference to the measured doppler shift.
...
There is no significant error...none at all for circular orbits.
Please explain why you think there is an error..
...
Yes, that's the error. The _published_ speed curve
will be based on the inverse period, the time
between pulse arrivals so that's what you need
to put into the simulation to make the curve
comparable.

George, the velocity will range from ~27000 +/-~0.01% m/s
Do you agree?

I am saying that, for any significant extinction
distance, the red line should have a greater
variation than the blue line. To find the true
speed, you adjust the velocity parameter until the
red line matches the published velocity curve. What
we need to sort out is why I think the red should
be higher than the blue.

I told you, the program deliberately normalises the heights of the two curves
to make shape comparison easier. If you like I will get it to plot a true
amplitude comparison.

They will also move closer and farther due to their
initially different speeds but that part will become
constant as the speeds equalise.

Yes..but their spacing overall will retain a periodic bunching.
It is not CONSTANT all the way along.

I think that's what I just said. It isn't constant
and reduces or grows until the speeds equalise
after which they remain unchanged regardless of
distance.

OK we agree on that.

Consider two pulses transmitted just before and just
after the neutron star passes behind the dwarf as seen
from Earth. This is the point of highest acceleration
and the second catches the first at the maximum rate.

First consider no extinction. The diagram shows the
earlier pulse 'a' already ahead of 'b' at the time
when b is emitted:

b a
b a
b a
*
a b
a b

The time between pulses goes to zero at the critical
distance. Now add extinction:

b a
b a
b a
b a
b a
b a
b a

The 'wavelength' settles down to a constant value but
it is less than the original.

George, George....
Consider what happens to pulses emitted when the pulsar is at the sides of the
orbit. ..where there is NO aceleration. They are also equally spaced for the
whole journey.

BUT THE SPACING IS NOT THE SAME AS THAT BETWEEN THE FORMER ONES a and b.

In other words, the normal doppler pattern is there whether you use BaTh or
constant c.
You have to include the difference in emission times of course.

Note that this effect
is in addition to the normal Doppler change due to
velocity alone (but at the location we are considering
the radial speed is zero).

Yes assume that is zero.

It is only that final pulse separation that we can
measure and which has been used to calculate the
27km/s value, and of course the published values assume
invariant speed. That means that if you want to compare
your program's output, specifically the blue line, with
published curves, you need to convert the received PRF
to a velocity _as_if_ the speed were always c, not
because of the physics but (if you like to think of it
this way) because that is the publishing convention.

In a nutshell, the shortened inter-pulse gap due to c+v
catch-up tricks us into thinking the orbital velocity
is higher than it really is. The red curve is the real
value and the blue curve is the "constant c" value
inferred from that shortened gap between pulses.

No you've got it all wrong George.
The BLUE curve is the actual one. (It will also be the one generated using pure
doppler very near the source).
The maximum ampitude of the red curve (from the doppler shifts the observer
measures) can never be higher than c+v.


Does that make it clearer Henry?

If you follow that, you should appreciate that instead
of saying the extinction is 6 light hours, you could
keep your 0.7 light year figure but drop the orbital
speed to 27 m/s. Of course that's not tenable for a
variety of other reasons but it might illustrate the
point, almost all the apparent "Doppler" shift would
actually be due to the pulse catch-up effect.

For those parameters, the red curve would be 27983 m/s
but the blue curve would be only 27 m/s, and because
most of the red curve is due to the acceleration at
the time of emission, there would be a 90 degree phase
difference.

You are not taking into account the effect of the delay in emission time. That
affects the spacing. It results in doppler wavelength shift for the constant c
model but not the BaTh one. For the latter, a doppler shift occurs during speed
change.

I think you should write a computer program to do all this George, instead of
trying to analyse the thing the way you are doing it.

It took me six years part-time to get it right....good luck.

George


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On Thu, 22 Feb 2007 13:20:06 +0000 (UTC), bz
wrote:

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

On Thu, 22 Feb 2007 02:48:38 +0000 (UTC), bz
wrote:

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

On Wed, 21 Feb 2007 15:33:43 +0000 (UTC), bz
wrote:

....
You ignore the effect that I have repeatedly mentioned, that the light
must come from the direction 'where the star was when the light was
emitted'[modified by aberration]. So WH variable stars with large
proper motion MUST result in the fast photons coming from a different
location in the sky than the slow photons. This would make the image
waltz back and forth in time with the orbit RATHER than showing up as
variations in brightness.

The main purpose of this thread was to ascertain whether or not there
was enough star movement to cause this kind of effect.
the general cinsensus is that most stars are too far away for this to
happen.

Of course. 'Most stars' are not even visible with the naked eye. Most
stars are in distant galaxies.

You like computer programs, enjoy, this one shows motion of stars.
http://www.rssd.esa.int/hipparcos/apps/ShowMotion.html

a google search for cepheid "with high proper motion"
turns up some very interesting hits.
http://adsabs.harvard.edu/abs/1979MNRAS.189..377P

well bob, most people here have convinced me that there is little
movement of the 'fixed satrs'


Fixed satrs! I like that
[quote]
Satres God of time and necessity. painted as an old man carrying a
sickle and an hourglass. same as the Roman god Saturn.
[unquote]

Amazing that 'most people here' could convince you of something that is NOT
true while many here have tried and failed to convince you of what is true.

We are not talking of 'fixed stars' when we speak of stars with high proper
motion.

Did you even go look at the animation at the link I posted?

Still too many questions left unanswered. I don't need the answers.

YOU are the one that needs the answers, if you are to ever be able to
support your theories. I have been trying to help you see that those
questions need to be answered.

I am producing the answer Bob.


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

The paranoid's delusions make him feel important.

A fool may ask more questions than 10 wise men can answer, but just because
someone asks a question does not mean that he is a fool. If it takes 20
wise men to answer the fools questions, he will know as much as 20 wise men
about the subject in question.--bz (c) 2007

"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

"Henri Wilson" HW@.... wrote in message
...
On Thu, 22 Feb 2007 10:17:35 -0000, "George Dishman"

wrote:


Most of our views on this are now in accord, I only
address the speed issue here and maybe pick up some
other minor points separately later.

First I'll take one paragraph from later on;


Your method doesn't take the effect of
the initial speed difference into account.

Don't be silly George, Of course it does. That's the whole basis of the
calculation.
The radial speed at each point around the orbit is c + vcos(A)

I said before you could treat cos(A) as being always 1.

A is a function of time George. I pointed that out way back.

I know.

I was thinking there of the angle between the line of
sight and the line between the barycentres. Your angle
is between to have used the line of sight and the
velocity which of course is essential but if you make
it the angle between the velocity and a line joining
the barycentres then there will be a negligible error,
essentially the view from infinity, and it will work
at zero distance to allow comparison with the conventional
model.

Thanks to Jeff Root for pointing out my misunderstanding
of your definition.

You are still completely misunderstanding the whole thing.

I did misunderstand which angle your were describing
previously. Consider this where B is the barycentre
of the system, P is the pulsar, D is the Dwarf and E
is the Earth:


C
B E
P

I was saying you could neglect the angle P-E-B. Obviously
you still need to take the direction of the pulsar velocity
into account.

"Henri Wilson" HW@.... wrote in message
. ..
On Wed, 21 Feb 2007 18:35:50 -0000, "George Dishman"
wrote:
...
I did explain Henry, at the critical distance the
gap between pulses is zero so your program should
report a value of c for the observed velocity curve
but the peak is the same height as the true value
which you entered as 0.0009. That's wrong by a
factor of 11000.

I think I know what you are trying to say here George.

At the critical distance, SOME pulses arrive together not ALL of them.
that is
because a cincave section of the orbit is such tat a large group of
pulses
will
arrive at a distant point over a very short time interval.
They will have started out with a range of speeds; that's why some catch
up
with the others.

Yes.

After extinction, they will all be traveling at about c wrt the source
BUT
their wavelengths will have changed so that their source speeds will
still
appear to be the correct ones, when measured with a grating at the
observer
distance..

No. We are not using a grating. Individual pulses have
their time of arrival noted against an atomic clock.
Remember they are 2.95 ms apart so the 'wavelength' is
885 km.

The inverse of the time between arrivals is the pulse
repetion frequency. That frequency is what is turned
into the published orbital parameters and is what give
the 339 Hz +/- 30 mHz values.

That's due to normal doppler 'bunching'.

No, it is what is observed. It results from a combination
of the normal bunching due to the varying distance from
Earth and also the catch-up effect.

BaTh bunching is virtually the same.

So my graph shows the 'no extinction' case...because I say extinction
makes no
difference to the measured doppler shift.
...
There is no significant error...none at all for circular orbits.
Please explain why you think there is an error..
...
Yes, that's the error. The _published_ speed curve
will be based on the inverse period, the time
between pulse arrivals so that's what you need
to put into the simulation to make the curve
comparable.

George, the velocity will range from ~27000 +/-~0.01% m/s
Do you agree?

I am saying that, for any significant extinction
distance, the red line should have a greater
variation than the blue line. To find the true
speed, you adjust the velocity parameter until the
red line matches the published velocity curve. What
we need to sort out is why I think the red should
be higher than the blue.

I told you, the program deliberately normalises the heights of the two
curves
to make shape comparison easier. If you like I will get it to plot a true
amplitude comparison.

That would help but what we need is the numerical
values.

They will also move closer and farther due to their
initially different speeds but that part will become
constant as the speeds equalise.

Yes..but their spacing overall will retain a periodic bunching.
It is not CONSTANT all the way along.

I think that's what I just said. It isn't constant
and reduces or grows until the speeds equalise
after which they remain unchanged regardless of
distance.

OK we agree on that.

Consider two pulses transmitted just before and just
after the neutron star passes behind the dwarf as seen
from Earth. This is the point of highest acceleration
and the second catches the first at the maximum rate.

First consider no extinction. The diagram shows the
earlier pulse 'a' already ahead of 'b' at the time
when b is emitted:

b a
b a
b a
*
a b
a b

The time between pulses goes to zero at the critical
distance. Now add extinction:

b a
b a
b a
b a
b a
b a
b a

The 'wavelength' settles down to a constant value but
it is less than the original.

George, George....
Consider what happens to pulses emitted when the pulsar is at the sides of
the
orbit. ..where there is NO aceleration. They are also equally spaced for
the
whole journey.

Yes, at those points you only get the velocity effect
but at any other location in the orbit the spacing is
affected by both the velocity _and_ the acceleration.

BUT THE SPACING IS NOT THE SAME AS THAT BETWEEN THE FORMER ONES a and b.

In other words, the normal doppler pattern is there whether you use BaTh
or
constant c.

The normal Doppler is there of course, I haven't disputed
that, but it isn't the whole story. The pulse spacing is
also affected by what I describe above and you need to take
that into account AS WELL to get the full answer.

You have to include the difference in emission times of course.

Yes.

Note that this effect
is in addition to the normal Doppler change due to
velocity alone (but at the location we are considering
the radial speed is zero).

Yes assume that is zero.

Only at that point, I agree with your description above
that velocity plays a part elsewhere.

It is only that final pulse separation that we can
measure and which has been used to calculate the
27km/s value, and of course the published values assume
invariant speed. That means that if you want to compare
your program's output, specifically the blue line, with
published curves, you need to convert the received PRF
to a velocity _as_if_ the speed were always c, not
because of the physics but (if you like to think of it
this way) because that is the publishing convention.

In a nutshell, the shortened inter-pulse gap due to c+v
catch-up tricks us into thinking the orbital velocity
is higher than it really is. The red curve is the real
value and the blue curve is the "constant c" value
inferred from that shortened gap between pulses.

No you've got it all wrong George.
The BLUE curve is the actual one.

Sorry, I warned you I might get the colours the wrong way
round.

(It will also be the one generated using pure
doppler very near the source).

Yes.

The maximum ampitude of the red curve (from the doppler shifts the
observer
measures) can never be higher than c+v.

No, it can be much higher because of later faster
pulses catching earlier slower ones. That effect
varies round the orbit of course but the highest
effect is at the point I described when the pulsar
is farthest from Earth.

Does that make it clearer Henry?

If you follow that, you should appreciate that instead
of saying the extinction is 6 light hours, you could
keep your 0.7 light year figure but drop the orbital
speed to 27 m/s. Of course that's not tenable for a
variety of other reasons but it might illustrate the
point, almost all the apparent "Doppler" shift would
actually be due to the pulse catch-up effect.

For those parameters, the red curve would be 27983 m/s
but the blue curve would be only 27 m/s, and because
most of the red curve is due to the acceleration at
the time of emission, there would be a 90 degree phase
difference.

You are not taking into account the effect of the delay in emission time.

I didn't mention it only because I assumed we were both
aware of it. The total change in pulse separation is of
course the combination of both effects.

That
affects the spacing. It results in doppler wavelength shift for the
constant c
model but not the BaTh one. For the latter, a doppler shift occurs during
speed
change.

I think you should write a computer program to do all this George, instead
of
trying to analyse the thing the way you are doing it.

It took me six years part-time to get it right....good luck.

Let's see what values you are getting, I hadn't realised
you had normalised the height of the peaks.

George

"Henri Wilson" HW@.... wrote in message
...
On Thu, 22 Feb 2007 11:23:57 -0000, "George Dishman"

wrote:


"Henri Wilson" HW@.... wrote in message
. ..
On Wed, 21 Feb 2007 22:34:29 -0000, "George Dishman"

wrote:
...
There you are Henry, we have derived an upper limit
on the extinction distance from the published data.
Now you understand what I was driving at, and
hopefully you also realise I really did understand
your model all along :-)

While some stars may have more or less dense plasma
around them, in general the distance should be around
that sort of level for all and shorter for stars with
a dense plasma. Note that it is much less than the
distance to the heliopause for the Sun.

George, I don't have a firm view as to why my distances are always
shorter
than the actual ones....but there must obviously be a simple
explanation.

The simple explanation is that SR is correct. From
your point of view though, as light passes through
a plasma we know it is affected and that could cause
some change to the speed. The obvious explanation
would be that absorption and re-emission at each
atom encountered immediately changes the speed to c
relative to that atom, but that would eliminate any
effects so your problem is why the extinction distance
isn't the mean path length.

George, this is the picture.
We have a neutron star rotating very rapidly and at the same time orbiting
a
dwarf star.
Some kind of radiation, presumeably magnetic, is emitted by the neutron
star.

Electromagnetic. Magnetic alone doesn't propagate.

We are assuming its speed wrt Earth varies between about c+/- 0.00009.

No, we are taking as a given that the time between
pulse arrivals varies by about 90 parts per million.
Some of that variation is due to the velocity but
some will be due to c+v pulses catching up to c-v
pulses a little in the time before extinction
equalises their speeds.

My theory says that for the pulses to be observed the way they are, there
must
be some kind of light speed unification taking place within one lightday
of the
system barycentre. Its speed approaches c in that time. Both 1.00009 c and
0.99991c become c.
If an inverse square law is involved, most of the change must occur in
much
less than 1 day.

It isn't inverse square, it is inverse exponential,
but either way most will be in that sort of time
frame.

This is perfectly in accordance with my concept of an EM FOR surrounding
large
mass centres. It is not a plain 'gravity' effect. That happens separately
and
shows up as Shapiro delay.

"Frame of reference" is a mathematical construct of
no relevance to the topic. Think of it as meaning a
coordinate system whose origin is the pulsar, nothing
more. Coordinates don't affect light.

The fact that so many brightness curves are reproducable using BaTh is
enough
to keep me convinced I'm right.

I think other factors are operating here.

There are no "other factors" in Ritzian theory to
operate aside from those already in your program.
You still need to fix that bug in the velocity
curve though.

There is no bug.

See my other post for details.

Circular orbits can appear slightly elliptical and vice versa.

Perhaps, but whether the distortion caused by variable
speed exactly eliminates that caused by Kepler's Second
Law is something you should show mathematically, and I
don't believe you can do that. As a result I think you
will find there remains a slight distortion even for
your best fit.

I can enlarge the curve and superimpose a sinewave on it.
I will do that just for you.

Instead, calculate the sine wave and then plot the
difference between the perfect sine wave and your
curve. That is the "residual" which you will find
in the published papers. Give the value for the
maximum of that curve.

"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."

Hmm but a genius in physics is unlikely to need to get
the dunces to integrate an exponential for him. Remember
your "challenge" that I solved in a few lines?

George you solved the wrong problem.

The integral is of hte form Total time= intgrl [1/(1+Ae^-kt)].dt
(c=1)

A solution is: t +log(1+Ae^-kt) between 0 and t.

I found a simple way to closely approximate the integral using the sum of
a GP
instead...it is also faster to run.

the terms are
1/(1+0.00009),1/(1+0.00009X),1/(1+0.00009X2),1/(1+0.00009X3).........1/(1+0.00009X^n)
Since the 0.00009 is small, this can be closely approximated with:
(1-0.00009),(1-0.00009X),(1-0.00009X2)...........................(1-0.00009X^n)
The sum is (n+1) -(1-0.00009)*(X^n-1)/(X-1) wherer n is the number of
light
days and X is the unification rate (eg., 0.99995 per Lday)

The single calculation t = vR/c^2 will be even quicker ;-)

That is really your biggest problem, you don't seem to
have the familiarity with maths that you need to follow
a lot of the arguments.

Now you're starting to sound like geesey....

You are still using an iterative method when a direct
calculation would do the job. It suggests you aren't
really comfortable with this level of maths.

At the moment you seem to be
struggling with the wavelength to velocity conversion
for your blue line for example.

I'M not....YOU are.

We'll see when you un-normalise the curves, I hadn't
realised you did that and thought you meant the physics
made their height the same.

George

On Fri, 23 Feb 2007 00:11:56 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Thu, 22 Feb 2007 11:23:57 -0000, "George Dishman"

wrote:

The simple explanation is that SR is correct. From
your point of view though, as light passes through
a plasma we know it is affected and that could cause
some change to the speed. The obvious explanation
would be that absorption and re-emission at each
atom encountered immediately changes the speed to c
relative to that atom, but that would eliminate any
effects so your problem is why the extinction distance
isn't the mean path length.

George, this is the picture.
We have a neutron star rotating very rapidly and at the same time orbiting
a
dwarf star.
Some kind of radiation, presumeably magnetic, is emitted by the neutron
star.

Electromagnetic. Magnetic alone doesn't propagate.

The pulsar is a rotating magnet. How the electric component comes into the
picture is not certain. It could easily be generated in radiation belts around
the whole binary system. That might act as a local EM reference frame and unify
the emitted light speeds.

We are assuming its speed wrt Earth varies between about c+/- 0.00009.

No, we are taking as a given that the time between
pulse arrivals varies by about 90 parts per million.
Some of that variation is due to the velocity but
some will be due to c+v pulses catching up to c-v
pulses a little in the time before extinction
equalises their speeds.

....and that results in exactly the same doppler shift as your own model.


My theory says that for the pulses to be observed the way they are, there
must
be some kind of light speed unification taking place within one lightday
of the
system barycentre. Its speed approaches c in that time. Both 1.00009 c and
0.99991c become c.
If an inverse square law is involved, most of the change must occur in
much
less than 1 day.

It isn't inverse square, it is inverse exponential,
but either way most will be in that sort of time
frame.

I was talking about whatever it is that causes the unification. I was
speculating that its effect must drop off with distance from the star....that's
over and above the exponential approach to equilibrium.


This is perfectly in accordance with my concept of an EM FOR surrounding
large
mass centres. It is not a plain 'gravity' effect. That happens separately
and
shows up as Shapiro delay.

"Frame of reference" is a mathematical construct of
no relevance to the topic. Think of it as meaning a
coordinate system whose origin is the pulsar, nothing
more. Coordinates don't affect light.

Not according to SR....
The origin of this frame is the barycentre of the pair.


Circular orbits can appear slightly elliptical and vice versa.

Perhaps, but whether the distortion caused by variable
speed exactly eliminates that caused by Kepler's Second
Law is something you should show mathematically, and I
don't believe you can do that. As a result I think you
will find there remains a slight distortion even for
your best fit.

I can enlarge the curve and superimpose a sinewave on it.
I will do that just for you.

Instead, calculate the sine wave and then plot the
difference between the perfect sine wave and your
curve. That is the "residual" which you will find
in the published papers. Give the value for the
maximum of that curve.

Why would I want to calculate it when the computer can do it for me and give
answers for a broad range of parameter values? You're not up to date George.



George you solved the wrong problem.

The integral is of hte form Total time= intgrl [1/(1+Ae^-kt)].dt
(c=1)

A solution is: t +log(1+Ae^-kt) between 0 and t.

I found a simple way to closely approximate the integral using the sum of
a GP
instead...it is also faster to run.

the terms are
1/(1+0.00009),1/(1+0.00009X),1/(1+0.00009X2),1/(1+0.00009X3).........1/(1+0.00009X^n)
Since the 0.00009 is small, this can be closely approximated with:
(1-0.00009),(1-0.00009X),(1-0.00009X2)...........................(1-0.00009X^n)
The sum is (n+1) -(1-0.00009)*(X^n-1)/(X-1) wherer n is the number of
light
days and X is the unification rate (eg., 0.99995 per Lday)

The single calculation t = vR/c^2 will be even quicker ;-)

That is really your biggest problem, you don't seem to
have the familiarity with maths that you need to follow
a lot of the arguments.

Now you're starting to sound like geesey....

You are still using an iterative method when a direct
calculation would do the job. It suggests you aren't
really comfortable with this level of maths.

George, I DO use an equation. ...the sum of the above GP.
The problem is, every sample point around the orbit has a different value for
v.

At the moment you seem to be
struggling with the wavelength to velocity conversion
for your blue line for example.

I'M not....YOU are.

We'll see when you un-normalise the curves, I hadn't
realised you did that and thought you meant the physics
made their height the same.

Their heights ARE almost the same for small magnitude variations.
Without extinction, the amplitude of the red curve cannot be any greater than
that of the blue one.

George


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On Thu, 22 Feb 2007 23:59:59 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Thu, 22 Feb 2007 10:17:35 -0000, "George Dishman"

Thanks to Jeff Root for pointing out my misunderstanding
of your definition.

You are still completely misunderstanding the whole thing.

I did misunderstand which angle your were describing
previously. Consider this where B is the barycentre
of the system, P is the pulsar, D is the Dwarf and E
is the Earth:


D
B E
P

I was saying you could neglect the angle P-E-B. Obviously
you still need to take the direction of the pulsar velocity
into account.

Yes OK.



No. We are not using a grating. Individual pulses have
their time of arrival noted against an atomic clock.
Remember they are 2.95 ms apart so the 'wavelength' is
885 km.

The inverse of the time between arrivals is the pulse
repetion frequency. That frequency is what is turned
into the published orbital parameters and is what give
the 339 Hz +/- 30 mHz values.

That's due to normal doppler 'bunching'.

No, it is what is observed. It results from a combination
of the normal bunching due to the varying distance from
Earth and also the catch-up effect.

There's NO 'catch up effect' in YOUR theory.

BaTh bunching is virtually the same.


I told you, the program deliberately normalises the heights of the two
curves
to make shape comparison easier. If you like I will get it to plot a true
amplitude comparison.

That would help but what we need is the numerical
values.

Why? So you can plot them just as the computer does?

I can print out the values if you like.

I told you that the red curve is an average speed of the light that arrives in
a set time interval. I htink you can imagine the effect bunching has on that.


OK we agree on that.

Consider two pulses transmitted just before and just
after the neutron star passes behind the dwarf as seen
from Earth. This is the point of highest acceleration
and the second catches the first at the maximum rate.

First consider no extinction. The diagram shows the
earlier pulse 'a' already ahead of 'b' at the time
when b is emitted:

b a
b a
b a
*
a b
a b

The time between pulses goes to zero at the critical
distance. Now add extinction:

b a
b a
b a
b a
b a
b a
b a

The 'wavelength' settles down to a constant value but
it is less than the original.

George, George....
Consider what happens to pulses emitted when the pulsar is at the sides of
the
orbit. ..where there is NO aceleration. They are also equally spaced for
the
whole journey.

Yes, at those points you only get the velocity effect
but at any other location in the orbit the spacing is
affected by both the velocity _and_ the acceleration.

BUT THE SPACING IS NOT THE SAME AS THAT BETWEEN THE FORMER ONES a and b.

In other words, the normal doppler pattern is there whether you use BaTh
or
constant c.

The normal Doppler is there of course, I haven't disputed
that, but it isn't the whole story. The pulse spacing is
also affected by what I describe above and you need to take
that into account AS WELL to get the full answer.

The program takes everything into account.
Why don't you experiment with it?

You have to include the difference in emission times of course.

Yes.

Note that this effect
is in addition to the normal Doppler change due to
velocity alone (but at the location we are considering
the radial speed is zero).

Yes assume that is zero.

Only at that point, I agree with your description above
that velocity plays a part elsewhere.

I'm becoming a bit confused as to what we are actually talking about now.

It is only that final pulse separation that we can
measure and which has been used to calculate the
27km/s value, and of course the published values assume
invariant speed. That means that if you want to compare
your program's output, specifically the blue line, with
published curves, you need to convert the received PRF
to a velocity _as_if_ the speed were always c, not
because of the physics but (if you like to think of it
this way) because that is the publishing convention.

In a nutshell, the shortened inter-pulse gap due to c+v
catch-up tricks us into thinking the orbital velocity
is higher than it really is. The red curve is the real
value and the blue curve is the "constant c" value
inferred from that shortened gap between pulses.

No you've got it all wrong George.
The BLUE curve is the actual one.

Sorry, I warned you I might get the colours the wrong way
round.

(It will also be the one generated using pure
doppler very near the source).

Yes.

The maximum ampitude of the red curve (from the doppler shifts the
observer
measures) can never be higher than c+v.

No, it can be much higher because of later faster
pulses catching earlier slower ones.

You are thinking of my green BRIGHTNESS curve. That's the one that graphs the
bunching of he pulsar pulses.
The blue and red ones are velocity curves...as measured using a grating.

That effect
varies round the orbit of course but the highest
effect is at the point I described when the pulsar
is farthest from Earth.

That's where maximum BUNCHING originates. (at infinity for a circular orbit)

Does that make it clearer Henry?

If you follow that, you should appreciate that instead
of saying the extinction is 6 light hours, you could
keep your 0.7 light year figure but drop the orbital
speed to 27 m/s. Of course that's not tenable for a
variety of other reasons but it might illustrate the
point, almost all the apparent "Doppler" shift would
actually be due to the pulse catch-up effect.

For those parameters, the red curve would be 27983 m/s
but the blue curve would be only 27 m/s, and because
most of the red curve is due to the acceleration at
the time of emission, there would be a 90 degree phase
difference.

You are not taking into account the effect of the delay in emission time.

I didn't mention it only because I assumed we were both
aware of it. The total change in pulse separation is of
course the combination of both effects.

The emission delay is very important. It is instrumental in determining the
phase relationship between brightness and velocity curves. It becomes quite
complicated for elliptical orbit because the yaw angle is also vital.

That
affects the spacing. It results in doppler wavelength shift for the
constant c
model but not the BaTh one. For the latter, a doppler shift occurs during
speed
change.

I think you should write a computer program to do all this George, instead
of
trying to analyse the thing the way you are doing it.

It took me six years part-time to get it right....good luck.

Let's see what values you are getting, I hadn't realised
you had normalised the height of the peaks.

But they are never far apart anyway.

George


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On 23 Feb, 09:07, HW@....(Henri Wilson) wrote:
On Fri, 23 Feb 2007 00:11:56 -0000, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
.. .
On Thu, 22 Feb 2007 11:23:57 -0000, "George Dishman"


Electromagnetic. Magnetic alone doesn't propagate.

The pulsar is a rotating magnet. How the electric component comes into the
picture is not certain.

Look up "near field", the effects significant in
antenna design and RFI problems. However, for the
pulsar the magnetic field just provides the energy
and some mechanism in the plasma probably produces
the actual radiation. The details aren't clear yet
AFAIK.

It could easily be generated in radiation belts around
the whole binary system. That might act as a local EM reference frame

You still haven't learned what "reference frame" means.

and unify
the emitted light speeds.

We are assuming its speed wrt Earth varies between about c+/- 0.00009.

No, we are taking as a given that the time between
pulse arrivals varies by about 90 parts per million.
Some of that variation is due to the velocity but
some will be due to c+v pulses catching up to c-v
pulses a little in the time before extinction
equalises their speeds.

...and that results in exactly the same doppler shift as your own model.

What do you mean by my "own model", SR or my
corrections to your Ritzian version?

It isn't inverse square, it is inverse exponential,
but either way most will be in that sort of time
frame.

I was talking about whatever it is that causes the unification. I was
speculating that its effect must drop off with distance from the star....that's
over and above the exponential approach to equilibrium.

OK.

This is perfectly in accordance with my concept of an EM FOR surrounding
large
mass centres. It is not a plain 'gravity' effect. That happens separately
and
shows up as Shapiro delay.

"Frame of reference" is a mathematical construct of
no relevance to the topic. Think of it as meaning a
coordinate system whose origin is the pulsar, nothing
more. Coordinates don't affect light.

Not according to SR....

Yes, in SR

The origin of this frame is the barycentre of the pair.

The origin of a frame is whatever origin you use
for the measured values.

Circular orbits can appear slightly elliptical and vice versa.

Perhaps, but whether the distortion caused by variable
speed exactly eliminates that caused by Kepler's Second
Law is something you should show mathematically, and I
don't believe you can do that. As a result I think you
will find there remains a slight distortion even for
your best fit.

I can enlarge the curve and superimpose a sinewave on it.
I will do that just for you.

Instead, calculate the sine wave and then plot the
difference between the perfect sine wave and your
curve. That is the "residual" which you will find
in the published papers. Give the value for the
maximum of that curve.

Why would I want to calculate it when the computer can do it for me ..

Oh Henry, obviously I meant you get your program
to do the calculation and add another curve to the
plots!

and give
answers for a broad range of parameter values? You're not up to date George.

George you solved the wrong problem.

The integral is of hte form Total time= intgrl [1/(1+Ae^-kt)].dt
(c=1)

A solution is: t +log(1+Ae^-kt) between 0 and t.

I found a simple way to closely approximate the integral using the sum of
a GP
instead...it is also faster to run.

the terms are
1/(1+0.00009),1/(1+0.00009X),1/(1+0.00009X2),1/(1+0.00009X3).........1/(1+0*.00009X^n)
Since the 0.00009 is small, this can be closely approximated with:
(1-0.00009),(1-0.00009X),(1-0.00009X2)...........................(1-0.00009*X^n)
The sum is (n+1) -(1-0.00009)*(X^n-1)/(X-1) wherer n is the number of
light
days and X is the unification rate (eg., 0.99995 per Lday)

The single calculation t = vR/c^2 will be even quicker ;-)

That is really your biggest problem, you don't seem to
have the familiarity with maths that you need to follow
a lot of the arguments.

Now you're starting to sound like geesey....

You are still using an iterative method when a direct
calculation would do the job. It suggests you aren't
really comfortable with this level of maths.

George, I DO use an equation. ...the sum of the above GP.
The problem is, every sample point around the orbit has a different value for
v.

I am suggesting you only need to calculate t = vR/c^2
for the value of v at each point rather than your
iterative sum at each point.

At the moment you seem to be
struggling with the wavelength to velocity conversion
for your blue line for example.

I'M not....YOU are.

We'll see when you un-normalise the curves, I hadn't
realised you did that and thought you meant the physics
made their height the same.

Their heights ARE almost the same for small magnitude variations.
Without extinction, the amplitude of the red curve cannot be any greater than
that of the blue one.

That is where you are wrong, without extinction the
red curve increases with distance until the peaks
reaches c at the critical distance. With extinction
the red curve starts rising above the blue but is
asymptotic to a constant curve and will be close to
that at several times the extinction distance.

George

On 23 Feb, 09:39, HW@....(Henri Wilson) wrote:
On Thu, 22 Feb 2007 23:59:59 -0000, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
.. .
On Thu, 22 Feb 2007 10:17:35 -0000, "George Dishman"
Thanks to Jeff Root for pointing out my misunderstanding
of your definition.

You are still completely misunderstanding the whole thing.

I did misunderstand which angle your were describing
previously. Consider this where B is the barycentre
of the system, P is the pulsar, D is the Dwarf and E
is the Earth:

D
B E
P

I was saying you could neglect the angle P-E-B. Obviously
you still need to take the direction of the pulsar velocity
into account.

Yes OK.

The benefit was just that you then get the conventional
value as a simple confidence check by setting the distance
to zero.

No. We are not using a grating. Individual pulses have
their time of arrival noted against an atomic clock.
Remember they are 2.95 ms apart so the 'wavelength' is
885 km.

The inverse of the time between arrivals is the pulse
repetion frequency. That frequency is what is turned
into the published orbital parameters and is what give
the 339 Hz +/- 30 mHz values.

That's due to normal doppler 'bunching'.

No, it is what is observed. It results from a combination
of the normal bunching due to the varying distance from
Earth and also the catch-up effect.

There's NO 'catch up effect' in YOUR theory.

I know Henry, the program is supposed to simulate
the physics of ballistic theory.

BaTh bunching is virtually the same.

I told you, the program deliberately normalises the heights of the two
curves
to make shape comparison easier. If you like I will get it to plot a true
amplitude comparison.

That would help but what we need is the numerical
values.

Why? So you can plot them just as the computer does?

I can print out the values if you like.

Just the value at the peak. The purpose is to allow
the parameters to be adjusted until this value reads
27km/s so we know when we have a match.

I told you that the red curve is an average speed of the light that arrives in
a set time interval. I htink you can imagine the effect bunching has on that.

I don't want to imagine, I want the comuter to
do that for me.

OK we agree on that.

Consider two pulses transmitted just before and just
after the neutron star passes behind the dwarf as seen
from Earth. This is the point of highest acceleration
and the second catches the first at the maximum rate.

First consider no extinction. The diagram shows the
earlier pulse 'a' already ahead of 'b' at the time
when b is emitted:

b a
b a
b a
*
a b
a b

The time between pulses goes to zero at the critical
distance. Now add extinction:

b a
b a
b a
b a
b a
b a
b a

The 'wavelength' settles down to a constant value but
it is less than the original.

George, George....
Consider what happens to pulses emitted when the pulsar is at the sides of
the
orbit. ..where there is NO aceleration. They are also equally spaced for
the
whole journey.

Yes, at those points you only get the velocity effect
but at any other location in the orbit the spacing is
affected by both the velocity _and_ the acceleration.

BUT THE SPACING IS NOT THE SAME AS THAT BETWEEN THE FORMER ONES a and b.

In other words, the normal doppler pattern is there whether you use BaTh
or
constant c.

The normal Doppler is there of course, I haven't disputed
that, but it isn't the whole story. The pulse spacing is
also affected by what I describe above and you need to take
that into account AS WELL to get the full answer.

The program takes everything into account.
Why don't you experiment with it?

I have, it gives the wrong value because you only take
the bunching due to c+v vs c-v into account on the
brightness curve, not the velocity curve. The predicted
velocity is derived from the time between pulses so you
need to take into account there too.

You have to include the difference in emission times of course.

Yes.

Note that this effect
is in addition to the normal Doppler change due to
velocity alone (but at the location we are considering
the radial speed is zero).

Yes assume that is zero.

Only at that point, I agree with your description above
that velocity plays a part elsewhere.

I'm becoming a bit confused as to what we are actually talking about now.

At any point arond the orbit, pulses are being
sent with a time gap of 2.95 ms. That gap is
reduced at the receiver for two reasons:

a) the velocity of the pulsar towards the receiver
means that consecutive pulses travel different
distances. That is the normal Doppler effect.

b) if the pulses are transmitted at different speeds
then faster pulses can 'catch up' to slower ones
reducing the gap (or 'fall behind' if the second
pulse is slower increasing the gap) and hence the
time between reception depends on how much of this
effect happens before extinction equalises the
speeds. This effect is not taken into account
in published velocity curves so the published
values will be higher or lower than the simple
Doppler value.

Part (a) is dependent on the radial component of velocity
at the time of transmission, part (b) depends on the
acceleration at the same time and of course both vary
round the orbit. Your program includes effect (a) but
not effect (b).

George

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

On Wed, 21 Feb 2007 15:37:12 +0000 (UTC), bz
wrote:

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



You must play by the rules of the game.
Everything must be consistent with c'=c+v. You must deal with all the
implications, you can not pick and choose which you want to deal with.

Rubbish

Rubbish?

How can you pick and choose effects while ignoring other predictable
effects and claim to be a follower of science, as describe it in your
book?

Bob, the only so called evidence AGAINST the BaTh was De Sitter's work.
We know now why that is wrong.

Incorrect. There is a LOT more evidence against BaTh.

http://www.mpe.mpg.de/ir/GC/index.php

[quote]
In 1953, however, Parry Moon and Domina Spencer analyzed a number of visual
binaries to see whether the phenomenon predicted by Bergmann would even be
visible in the first place.7 They assumed the Ritz hypothesis8, but their
computations showed that Bergmann's predicted multiple images for binaries
would not, in fact, be observed. (They do not elaborate on de Sitter's
prediction of spurious eccentricities, and they do not mention whether they
reexamined his data or not.) Hence, they concluded that visual binaries
proved absolutely nothing about the constancy of the velocity of light.

[emphasis mine]In the same article, Moon and Spencer performed a similar
analysis of spectroscopic binaries and of Cepheid variables.9 They
concluded that the Ritz hypothesis would produce spurious spectral lines,
but no such phenomenon was observed.[end emphasis]
.....
is Fox's criticism-that the observations of de Sitter and Bergmann did not
take the Ewald and Oseen extinction effect into account-still valid?
Definitely not, for by 1964 direct evidence for the validity of Einstein's
postulate on the velocity of light was provided by a number of
experimenters: D. Sadeh; T.A. Filippas and J.G. Fox; and T. Alvager et
al.14 All of these experimenters measured the velocity of gamma rays which
had been emitted by decaying subatomic particles moving at nearly the speed
of light. In every case, the velocity of the gamma rays equaled that of the
normal velocity of light in free space. In no case did the velocity of the
gamma rays behave as proposed by Ritz.

In addition to the above Earth-based experiments, in 1977 K. Brecher used
radiation from pulsars (rotating neutron stars which emit radiation in a
periodic manner) to show that the speed of light was independent of the
motion of the source.15 Neither Brecher's experiment nor the ones mentioned
in the preceding paragraph were subject to Fox's criticism. Hence,
observations of both terrestrial and extra-terrestrial phenomena have shown
once and for all that Ritz's hypothesis is invalid. .... [unquote]





--
bz

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




--
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