"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