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