Why are the 'Fixed Stars' so FIXED?

On 15 Mar 2007 19:17:29 -0700, "Jerry" wrote:

On Mar 15, 6:00 pm, HW@....(Henri Wilson) wrote:

ALL THE VELOCITY CURVES EVER PRODUCED USING SPECTRAL DOPPLER
SHIFTS ARE PROBABLY VERY WRONG.

That's your way of hiding from the fact that your program fails
to produce the correct velocity curves.

Utterly pathetic.

You will never make it in medicine with your negative attitude problem....


Jerry


"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 Mar 15, 5:32 pm, HW@....(Henri Wilson) wrote:

I also had to adjust both the comparative brightness and
orbit speed of the 'outer star'. Both values are about 0.4
of the inner star. ...which provides an indication of the
relative masses. I achieved an even closer match when I
included a third object wirth a 90 degree phase shift.

YOU CAN'T DO THAT!!!

There is no way that adding the third object where you do
could result in a stable orbital configuration.

In every paper I have read about cepheids the authors
state straight out that they have no model that can
explain the brightness variations .

You are obviously not up on the latest research. With
recent advances in supercomputer capabilities, it has been
possible for astrophysicists to include in their models
effects that previously had to be ignored, because modeling
those effects required computational power far exceeding
that which had been available.

The state of the art in Cepheid modeling as of early 2005
is described in the following link:
http://www.lesia.obspm.fr/astro/cepheids/program.html

In the last couple of years, I've seen even better results!

Jerry

On 16 Mar, 02:33, HW@....(Henri Wilson) wrote:
On Fri, 16 Mar 2007 00:18:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message
.. .
On 15 Mar 2007 01:26:03 -0700, "George Dishman"

Nope, that would be of no use at all. Consider a simple
circular orbit of the pulsar P around the barycentre B
as seen by an observer O very far away (not to scale):

B

P x P' O

| |
|- D -|

Light from the two locations P and P' would be launched
with the same speed towards the observer, c' = c+0.7v,
because I've drawn them at 45 degrees to the LoS. The
light from P would be expected to take D/c' longer to
reach us.

That's a strange drawing George.
The barycentre should be where x is.
Anyway I know what you mean.

I don't think you do. Let me add the location of the
companion dwarf as C and draw the two situations
separately. Here's the first:

C

B

P O

The companion is lighter so it's farther from B. Here
is the second situation quarter of an orbit later:

C

B

P O

Of course the observer sholud be farr off to the
right of your screen

Oh all right. I thought you were trying to draw something else.
Same result anyway.
....
Point x is midway between P and P' where the light path
is perpendicular to line x-B. In ballistic theory the
gravity of the star accelerates the light between P and
x and then slows it between x and P' so that the speed
at P' is the same as light emitted at P'. Everything
from there to O is the same. The time it takes the light
to get from P to P' is therefore slightly _less_ than
D/c' because the mean speed is slightly higher than c'.
The Shapiro effect is the difference between that time
and D/c'.

Yes I'm aware of this. The average speed is faster than c' between P and
P'.

Right so the signal arrives earlier, it is not a delay.
The gravitational redshift is identical in each case as
is the eventual speed.

that's right.

OK, now we have cleared that up, if you plot the
alteration of arrival time as a function of the
phase, you will find it peaks when the source is
behins the companion and the relative width of the
peak depends on the inclination of the orbit. There
will be no effect for face on and a high narrow peak
for nearly edge on. However, there will be almost no
velocity effect since any increase between P and x
is always matched by a corresponding decrease between
x and P'. That's why we can use it as a reference for
the phase.

Consider a pulsar in an edge-on circular orbit.

Pulses from the near and far sections of the orbit move towards you at c
and that from the edges at c+v and c-v.

Bunching of pulses is a maximum at maximum acceleration, ie., for pulses
emitted from the far section of the orbit. It is minimum for those emitted
at the near, or 'convex' section.

There is also the effect that consecutive pulses from the
edge travel slightly differnt distances to reach us. Those
from the edge where the source is approaching us travel
progressively shorter distances so are slightly bunched by
the velocity while those on th other side travel a little
longer each time so are moved apart.

However, the 'bunched section' moves towards the observer at a slower
speed than does the group of pulses from the edges.

Slower than those from one edge, faster than from the other.

Now, my original method does not take this into account, although the red
velocity curve it generates actually shows the arrival velocities.

The velocities affect the 'y' position but the changed
time of arrival affects the 'x' position. However, any
change in that from one pulse to the next also affects
the _relative_ separation hence looks like a modification
to the velocity.

I cannot yet see how your 'pulses separation' method does NOT include the
VDoppler.

It SHOULD include it but if as you say above your program
does not take this into account, that could explain why it
is missing the VDoppler effect.

I'm working on it. I'll eventually find what's happening.
....
It will be done....but it isn't as simple as one would think.

I can see that your approach might make it tricky. Let
me know when you crack it.

George

On 15 Mar 2007 19:43:54 -0700, "Jerry" wrote:

On Mar 15, 5:32 pm, HW@....(Henri Wilson) wrote:

I also had to adjust both the comparative brightness and
orbit speed of the 'outer star'. Both values are about 0.4
of the inner star. ...which provides an indication of the
relative masses. I achieved an even closer match when I
included a third object wirth a 90 degree phase shift.

YOU CAN'T DO THAT!!!

There is no way that adding the third object where you do
could result in a stable orbital configuration.

In every paper I have read about cepheids the authors
state straight out that they have no model that can
explain the brightness variations .

You are obviously not up on the latest research. With
recent advances in supercomputer capabilities, it has been
possible for astrophysicists to include in their models
effects that previously had to be ignored, because modeling
those effects required computational power far exceeding
that which had been available.

The state of the art in Cepheid modeling as of early 2005
is described in the following link:
http://www.lesia.obspm.fr/astro/cepheids/program.html

In the last couple of years, I've seen even better results!

See http://www.users.bigpond.com/hewn/bunching.jpg
.....then burn all yer books...

Jerry


"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 16 Mar 2007 00:53:29 -0700, "George Dishman"
wrote:

On 16 Mar, 02:33, HW@....(Henri Wilson) wrote:
On Fri, 16 Mar 2007 00:18:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message
.. .

It SHOULD include it but if as you say above your program
does not take this into account, that could explain why it
is missing the VDoppler effect.

I'm working on it. I'll eventually find what's happening.
...
It will be done....but it isn't as simple as one would think.

I can see that your approach might make it tricky. Let
me know when you crack it.

George

See http://www.users.bigpond.com/hewn/bunching.jpg

This shows how the phase of the TRUE velocity maximum changes with distance
when compared with the maximum bunching. The latter (maximum pulse arrival
rate) is wrongly assumed to indicate the maximum doppler shift.

I think this answers your questions George.

The standard doppler approach applied to pulse arrival rate gives a completely
wrong answer.


"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 Mar 16, 3:23 am, HW@....(Henri Wilson) wrote:
On 15 Mar 2007 19:43:54 -0700, "Jerry" wrote:

You are obviously not up on the latest research. With
recent advances in supercomputer capabilities, it has been
possible for astrophysicists to include in their models
effects that previously had to be ignored, because modeling
those effects required computational power far exceeding
that which had been available.

The state of the art in Cepheid modeling as of early 2005
is described in the following link:
http://www.lesia.obspm.fr/astro/cepheids/program.html

In the last couple of years, I've seen even better results!

Seehttp://www.users.bigpond.com/hewn/bunching.jpg
....then burn all yer books...

All you have in the above link is a bunch of cartoons.

Show me that you can simultaneously match the luminosity
and radial velocity curves of RT Aurigae.

Thus far, you have done nothing but emit a lot of hot
air.

Your version of huff-puff, I suppose.

Jerry

"Jerry" wrote in message ups.com...
On Mar 16, 3:23 am, HW@....(Henri Wilson) wrote:
On 15 Mar 2007 19:43:54 -0700, "Jerry" wrote:

You are obviously not up on the latest research. With
recent advances in supercomputer capabilities, it has been
possible for astrophysicists to include in their models
effects that previously had to be ignored, because modeling
those effects required computational power far exceeding
that which had been available.

The state of the art in Cepheid modeling as of early 2005
is described in the following link:
http://www.lesia.obspm.fr/astro/cepheids/program.html

In the last couple of years, I've seen even better results!

Seehttp://www.users.bigpond.com/hewn/bunching.jpg
....then burn all yer books...

All you have in the above link is a bunch of cartoons.

Show me that you can simultaneously match the luminosity
and radial velocity curves of RT Aurigae.

Thus far, you have done nothing but emit a lot of hot
air.

HAHAHA!
Pot. Kettle. Black.

On 16 Mar, 09:32, HW@....(Henri Wilson) wrote:
On 16 Mar 2007 00:53:29 -0700, "George Dishman" wrote:
On 16 Mar, 02:33, HW@....(Henri Wilson) wrote:
On Fri, 16 Mar 2007 00:18:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message
.. .
It SHOULD include it but if as you say above your program
does not take this into account, that could explain why it
is missing the VDoppler effect.

I'm working on it. I'll eventually find what's happening.
...
It will be done....but it isn't as simple as one would think.

I can see that your approach might make it tricky. Let
me know when you crack it.

See http://www.users.bigpond.com/hewn/bunching.jpg

This shows how the phase of the TRUE velocity maximum changes with distance
when compared with the maximum bunching. The latter (maximum pulse arrival
rate) is wrongly assumed to indicate the maximum doppler shift.

It is hard to tell but it looks as though the
pulses in the top line are regularly spaced. If
so it is wrong, they should be bunched closest
at the 90 degree mark because each one travels
less distance to the observer than the previous
pulse (speed is constant at c+v) and most widely
spaced at the 270 degree mark where each travels
farther (at c-v). The bottom line looks right in
that the acceleration dominates and the maximum
is at the 0 degree mark (which is also where the
Shapiro delay peaks).

I think this answers your questions George.

No, my first question was to find the row where
the combination of the velocity and acceleration
effects gives a maximum bunching at 45 degrees
and you haven't shown that line.

Specifically though you need that incorporated
into your program so that we can find the distance
at which the measured phase shift occurs and you
may then want to also consider elliptical orbits
which will change both velocity and acceleration
versus phase and yaw due to Kepler's laws. The
diagram has been helpful as it shows you are still
not taking the velocity effect into account and
hopefully will give you a steer on how to do that,
but it is only a step to fixing the error in your
program.

Once you do that, you need to be able to say what
extinction distance produces a phase shift of the
order of 10^-5 degrees so you need to be able to
get numbers out of it, but we can estimate it from
the 45 degree figure with a bit of simple trig.

George

"Henri Wilson" HW@.... wrote in message
...
On Fri, 16 Mar 2007 00:22:44 -0000, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
news
On Wed, 14 Mar 2007 23:47:58 -0000, "George Dishman"
wrote:

I also had to adjust both the comparative brightness and orbit speed of
the
'outer star'. Both values are about 0.4 of the inner star. ...which
provides an
indication of the relative masses. I achieved an even closer match when
I
included a third object wirth a 90 degree phase shift.

You can't do that, it's an unstable configuration. You
could get away with one at a Lagrange point but there
is a limit on the mass ratios.

I wasn't suggesting that an object was in orbit 90 out. As far as we know
that
is indeed impossible.

Fine, so you are not allowed to put one into your simulation
and claim you have succeeded. In fact you told me you got
the motion of the stars by simulating Newtonian gravity so
your configuration should have been unstable. It looks as
though that part is buggy too.

...but there could be other reasons...tidal effects(?)

Then simulate tidal effects. All you can do for now is use
two stars and get the best fit. If the residuals are within
the observational uncertainty you have a match and if not
you don't.

The fact that it was 90 and not 80 or 100 made me wonder.

Pointless since it cannot exist.

I was wondering about the material that is falling into the neutron star.
If it is spinning, its speed would drop of with distance. If it wasn't
spinning
the pulsar would be slowing down.

So look up the rate of change of the pulsar frequency, it is
one of the key published values.

Of course, you can create any possible shape with sufficient
harmonics but Keplerian orbits produce limits, that is the
anture of the test. You can't just add more factors.

Everything I add is strictly in accordance with the BaTh. I cannot
simply add
any old curve to produce the one I want. There are strict limitations
particularly for elliptical orbits.

Yes, and a third object is not allowed !

Of course it is....many star curves clearly involve a third or more
object.

Then those curves will almost certainly be failures too, you
cannot have a stable configuration with a third object except
under _very_ limited conditions (e.g. figure of eight or the
very disparate separations like the Sirius system).

I don't think you have fully realised the complexity of this whole issue
George.

I don't think you realise the constraints Keplerian orbits
place on you Henry.

George

On 16 Mar 2007 03:19:00 -0700, "George Dishman"
wrote:

On 16 Mar, 09:32, HW@....(Henri Wilson) wrote:
On 16 Mar 2007 00:53:29 -0700, "George Dishman" wrote:
On 16 Mar, 02:33, HW@....(Henri Wilson) wrote:
On Fri, 16 Mar 2007 00:18:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message
.. .
It SHOULD include it but if as you say above your program
does not take this into account, that could explain why it
is missing the VDoppler effect.

I'm working on it. I'll eventually find what's happening.
...
It will be done....but it isn't as simple as one would think.

I can see that your approach might make it tricky. Let
me know when you crack it.

See http://www.users.bigpond.com/hewn/bunching.jpg

This shows how the phase of the TRUE velocity maximum changes with distance
when compared with the maximum bunching. The latter (maximum pulse arrival
rate) is wrongly assumed to indicate the maximum doppler shift.

It is hard to tell but it looks as though the
pulses in the top line are regularly spaced.

That line is the layout immediately after emission (one orbit) All the pulses
are (almost) equally spaced.

If
so it is wrong, they should be bunched closest
at the 90 degree mark because each one travels
less distance to the observer than the previous
pulse (speed is constant at c+v) and most widely
spaced at the 270 degree mark where each travels
farther (at c-v).

George, each line represents a distance further from the source. The diagram is
not wrong.

The maximum bunching occurs at the 0/360 mark. (furthest from observer).
Widest spacing is at the 180 mark. That's how ADoppler works.

The bottom line looks right in
that the acceleration dominates and the maximum
is at the 0 degree mark (which is also where the
Shapiro delay peaks).

All the lines are correct. They show the layout at different distances,
increasing down the screen..

The program merely shows how each pulse moves after emission, given that its
velocity is c+vsin(x/T).


I think this answers your questions George.

No, my first question was to find the row where
the combination of the velocity and acceleration
effects gives a maximum bunching at 45 degrees
and you haven't shown that line.

The acceleration term dominates from the start.
The source velocity is so small that the pulses are virtually evenly spaced
after one orbit. Do you see that? That is represented by the top line....with
the first pulses at the RHS.

As distance increases, the pulses emitted at the 90 mark move towards the
leading ones, causing bunching there.




Specifically though you need that incorporated
into your program so that we can find the distance
at which the measured phase shift occurs and you
may then want to also consider elliptical orbits
which will change both velocity and acceleration
versus phase and yaw due to Kepler's laws. The
diagram has been helpful as it shows you are still
not taking the velocity effect into account and
hopefully will give you a steer on how to do that,
but it is only a step to fixing the error in your
program.

George you are totally confused.
The diagram is correct.

Astronomers unwittingly use the pulse arrival rate as a measure of doppler
shift.
They believe the higher the arrival rate, the faster the radial velocity
towards Earth.

As you can see, this is is completely wrong, both in magnitude and phase wrt
the brightness curve. The true maximum radial velocity occurs at the 90 degree
mark.

Once you do that, you need to be able to say what
extinction distance produces a phase shift of the
order of 10^-5 degrees so you need to be able to
get numbers out of it, but we can estimate it from
the 45 degree figure with a bit of simple trig.

George you don't seem to understand. I suggest you write your own program.
It is quite simple really...for circular orbits.

Set up an array of speeds around the orbit:

For K=0 to 360
lightspeed(K) = 2000 * (Sin(pi/180 * K) * vone)
next

(the starting point is as shown in my diagram.)

then:
using a timer, repeat the loop:

For j = 0 To 90 (gives 90 pulse)
If j = 22 Or j = 45 Or j = 67 Then n = 0 Else n = 255 (red line showing
90,180, 270 points)

Line ((140 * j) + (sec * lightspeed(360 - (4 * j))), 2000)-((140 * j) + (sec *
lightspeed(360 - (4 * j))), 2200), RGB(255, n, 0)
Next

sec= sec+1


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.