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Old March 29th 07, 08:09 AM posted to sci.physics.relativity,sci.astro
George Dishman[_1_]
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Default Why are the 'Fixed Stars' so FIXED?

On 29 Mar, 01:32, HW@....(Henri Wilson) wrote:
On 28 Mar 2007 06:27:41 -0700, "George Dishman" wrote:
On 28 Mar, 11:40, HW@....(Henri Wilson) wrote:
On 28 Mar 2007 02:16:59 -0700, "George Dishman" wrote:
On 28 Mar, 08:10, HW@....(Henri Wilson) wrote:


The diagram would be like this:


g h --- O


+
B


The pulsar sends one pulse from g and the next from h,
it is orbiting round the barycentre B and the observer
is at O. Obviously there is a v*cos(theta) term for
other parts of the orbit, it is the distance change
in the direction of the line of sight that matters.


I have
incorporated that by adding an Rsin(x) term to the star distance. It is
generally negligible.


It will certainly be small but it is not negligible, it
will produce a 45 degree phase shift when the ADoppler
is about 93 parts per million too and in fact we know
that the VDoppler is probably larger than the ADoppler
_except_that_ the phase can be changed by the effect you
describe at the top of the post regarding an elliptical
orbit looking circular.


I think I had it right before.
The distance for 45 deg phase difference is about 0.0007 LY.
It is independent of velocity.


OK, that is the sort of value I would expect. Now
the general gist of my argument is this: you get
a 45 degree phase shift at 0.0007 LY so you would
expect to get of the order of 5 degrees at a 1/10th
of that distance where the ADoppler only adds a
small fraction to the VDoppler.

You made the point that an elliptical orbit could
look circular provided the periastron was on the
line of sight because the distortion of the sine
wave from the variable speed is cancelled by the
distortion caused by the c+v effect.

A slight change in your yaw factor could then
change the relative phase of those factors to
give a net phase change of a few degrees. That
could cancel the phase shift due to ADoppler
and again make the orbit look circular.

The bottom line then is that knowing we see what
looks like a circular orbit (or at least very low
eccentricity) there is a relationship between the
extinction distance, the true eccentricity and the
yaw.

From your other reply:


That is what I was alluding to a couple of weeks ago. For
small values you can probably get a match by eye but the
equation for an ellipse and those for Kepler's Laws are
quite different from the effect of ballistic theory. It
would be a curious though unimportant coincidence if they
exactly matched. Just as Ptolemy was able to get a good but
imperfect match with combined circles, I think if you did
the analytical investigation, you would find there was a
small difference but perhaps third or fourth order. That
is what would show up as the shape of a pattern in your
residuals.


I think it is quite likely that there is an exact match. It isn't unreasonable.


Given the form of the equations, I disagree but if
you do the calculation, you might prove me wrong.

For different eccentricitiers, the curve becomes a sinewave at diffferent
distances (for he same maximum velocity)


Possibly, but I think the ADoppler distortion continues
to increase with distance and eventually causes multiple
images while the Keplerian distortion will be asymptotic
to some curve as the yaw approaches 90 degrees. The
question is how much the cancellation degrades as higher
order terms become more important. Your simulation is
the easiest way to investigate that.

The end result should be an upper limit on the speed
equalisation distance based on the uncertainty in the
orbital phase and the eccentricity.

George