Why are the 'Fixed Stars' so FIXED?

On 20 Feb 2007 02:18:17 -0800, "George Dishman"
wrote:

On 19 Feb, 23:17, HW@....(Henri Wilson) wrote:
On 18 Feb 2007 19:20:31 -0800, "Leonard Kellogg" wrote:
Henri Wilson wrote:
...
I use symbolic pulses from a star of constant brightness
emitted at equi-temporal points around the orbit. These
travel at varying c+cos(v) speeds towards a distant obsever.
The rate at which they arrive at the observer should then
simulate its brightness curve there.

So apply that to the pulsar.

There is absolutely no point....unless you can provide a reliable curve showing
the variation in arrival rate of the pulses over time.

Henry, I have already done that several times. In
round figures the PRF is 339 Hz and that is varied
by +/- 30.5 mHz. The exact numbers are in my previous
posts.

The velocity curve that would be published is just
a sine wave (near zero eccentricity) with an
amplitude of c * 0.0305 / 339 = 27983 m/s.

That should be the same
as my 'brightness curve'. I can't make sense of the curve published by Jacoby
et al

Also, I cannot adjust the number of pulses I sample per orbit (122 million in
this case) without changing the code a bit. I can do it but it will take a
little time

You don't need to, it is only the frequency ratio
that matters. You do need to fix the velocity curve
calculation though, I'll explain that in reply to
your post giving the details of your current
calculation.

OK
The frequency variation correspondes to a magnitude change of about 0.2.
(CMIIW)

have look at: http://www.users.bigpond.com/hewn/J1909-3744a.jpg

Here the magnitude change is 1 (too high) but I included it because it shows
how the OBSERVED sine velocity curve corresponds to an orbit with ~0.15
eccentricity with Yaw 90. (periastron furthest from observer). Distance = 4 LY
for this mag change.

and: http://www.users.bigpond.com/hewn/J1909-3744b.jpg
mag change ~0.2
Period = 0.0042 years
max velocity=0.0000933c.

To obtain curve b, I have to plug in a distance of less than 1 LY....more like
0.7 LYs.
This order of 'extinction length' is quite consistent with those I have derived
from short period contact binaries.

In curve b, the magnitude change is smaller and a sine-like red velocity curve
corresponds with an e ~ 0.06, yaw -90.
A circular orbit results in a clearly skewed red curve.
So my theory says the orbit is NOT circular at all.

provide me with a good curve of pulse arrival times and i can probably do what
you ask.

Already done, repeatedly.





George

"Henri Wilson" HW@.... wrote in message
...
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"Henri Wilson" HW@.... wrote in message
...
On 20 Feb 2007 02:18:17 -0800, "George Dishman"
wrote:

Henry, I have already done that several times. In
round figures the PRF is 339 Hz and that is varied
by +/- 30.5 mHz. The exact numbers are in my previous
posts.

The velocity curve that would be published is just
a sine wave (near zero eccentricity) with an
amplitude of c * 0.0305 / 339 = 27983 m/s.

That should be the same
as my 'brightness curve'. I can't make sense of the curve published by
Jacoby
et al
....
OK
The frequency variation correspondes to a magnitude change of about 0.2.
(CMIIW)

I think you used Hz instead of mHz, it is a brightness
ratio of 1.00018 which corresponds to a magnitude change
of about 0.0001955. Your approach is right though.

and: http://www.users.bigpond.com/hewn/J1909-3744b.jpg
mag change ~0.2
Period = 0.0042 years
max velocity=0.0000933c.

To obtain curve b, I have to plug in a distance of less than 1 LY....more
like
0.7 LYs.
This order of 'extinction length' is quite consistent with those I have
derived
from short period contact binaries.

In curve b, the magnitude change is smaller and a sine-like red velocity
curve
corresponds with an e ~ 0.06, yaw -90.
A circular orbit results in a clearly skewed red curve.
So my theory says the orbit is NOT circular at all.

OK, that's exactly the sort of difference in prediction I have
been interested in. Pop in the right brightness and let's see
what you get now. My guess is your 'extinction length' will
need to go down by a factor of 1000 to 0.0007 light years or
about 6 light hours !!!!

George

On Wed, 21 Feb 2007 10:59:41 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On 20 Feb 2007 02:18:17 -0800, "George Dishman"
wrote:

Henry, I have already done that several times. In
round figures the PRF is 339 Hz and that is varied
by +/- 30.5 mHz. The exact numbers are in my previous
posts.

The velocity curve that would be published is just
a sine wave (near zero eccentricity) with an
amplitude of c * 0.0305 / 339 = 27983 m/s.

That should be the same
as my 'brightness curve'. I can't make sense of the curve published by
Jacoby
et al
...
OK
The frequency variation correspondes to a magnitude change of about 0.2.
(CMIIW)

I think you used Hz instead of mHz, it is a brightness
ratio of 1.00018 which corresponds to a magnitude change
of about 0.0001955. Your approach is right though.

Yes i did use hz onstead of mhz. Sory about that.

and: http://www.users.bigpond.com/hewn/J1909-3744b.jpg
mag change ~0.2
Period = 0.0042 years
max velocity=0.0000933c.

To obtain curve b, I have to plug in a distance of less than 1 LY....more
like
0.7 LYs.
This order of 'extinction length' is quite consistent with those I have
derived
from short period contact binaries.

In curve b, the magnitude change is smaller and a sine-like red velocity
curve
corresponds with an e ~ 0.06, yaw -90.
A circular orbit results in a clearly skewed red curve.
So my theory says the orbit is NOT circular at all.

OK, that's exactly the sort of difference in prediction I have
been interested in. Pop in the right brightness and let's see
what you get now. My guess is your 'extinction length' will
need to go down by a factor of 1000 to 0.0007 light years or
about 6 light hours !!!!

I cannot give you an exact figure because of way magnitude is calculated...but
it is less than 1 lightday.

.....Not imposible since it is a neutron star.

I think other factors are operating here.




George

"Henri Wilson" HW@.... wrote in message
...
On Wed, 21 Feb 2007 10:59:41 -0000, "George Dishman"

wrote:

The frequency variation correspondes to a magnitude change of about 0.2.
(CMIIW)

I think you used Hz instead of mHz, it is a brightness
ratio of 1.00018 which corresponds to a magnitude change
of about 0.0001955. Your approach is right though.

Yes i did use hz onstead of mhz. Sory about that.

No problem, easily done.

and: http://www.users.bigpond.com/hewn/J1909-3744b.jpg
mag change ~0.2
Period = 0.0042 years
max velocity=0.0000933c.

To obtain curve b, I have to plug in a distance of less than 1
LY....more
like
0.7 LYs.
This order of 'extinction length' is quite consistent with those I have
derived
from short period contact binaries.

In curve b, the magnitude change is smaller and a sine-like red velocity
curve
corresponds with an e ~ 0.06, yaw -90.
A circular orbit results in a clearly skewed red curve.
So my theory says the orbit is NOT circular at all.

OK, that's exactly the sort of difference in prediction I have
been interested in. Pop in the right brightness and let's see
what you get now. My guess is your 'extinction length' will
need to go down by a factor of 1000 to 0.0007 light years or
about 6 light hours !!!!

I cannot give you an exact figure because of way magnitude is
calculated...but
it is less than 1 lightday.

....Not imposible since it is a neutron star.

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.

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.

George

On Wed, 21 Feb 2007 22:34:29 -0000, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Wed, 21 Feb 2007 10:59:41 -0000, "George Dishman"

wrote:

The frequency variation correspondes to a magnitude change of about 0.2.
(CMIIW)

I think you used Hz instead of mHz, it is a brightness
ratio of 1.00018 which corresponds to a magnitude change
of about 0.0001955. Your approach is right though.

Yes i did use hz onstead of mhz. Sory about that.

No problem, easily done.

and: http://www.users.bigpond.com/hewn/J1909-3744b.jpg
mag change ~0.2
Period = 0.0042 years
max velocity=0.0000933c.

To obtain curve b, I have to plug in a distance of less than 1
LY....more
like
0.7 LYs.
This order of 'extinction length' is quite consistent with those I have
derived
from short period contact binaries.

In curve b, the magnitude change is smaller and a sine-like red velocity
curve
corresponds with an e ~ 0.06, yaw -90.
A circular orbit results in a clearly skewed red curve.
So my theory says the orbit is NOT circular at all.

OK, that's exactly the sort of difference in prediction I have
been interested in. Pop in the right brightness and let's see
what you get now. My guess is your 'extinction length' will
need to go down by a factor of 1000 to 0.0007 light years or
about 6 light hours !!!!

I cannot give you an exact figure because of way magnitude is
calculated...but
it is less than 1 lightday.

....Not imposible since it is a neutron star.

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 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. Circular orbits can appear slightly elliptical and vice versa.

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.

"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.

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.

"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?

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. At the moment you seem to be
struggling with the wavelength to velocity conversion
for your blue line for example. Anyway, see if my latest
attempt to explain it lets the penny drop and we'll see
where that takes your program.

George

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.