Why are the 'Fixed Stars' so FIXED?

On 28 Mar, 08:10, HW@....(Henri Wilson) wrote:
On Sun, 25 Mar 2007 23:34:03 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message

George, when you can, have a look at

http://www.users.bigpond.com/hewn/ellip_circle.jpg

This shows how an elliptical orbit can produce a near perfect sine wave under
certain condition whilst the circular orbit produces nothing like one for
exactly the same parameter values.

Yaw angle is -90 (periastron closest to observer).

The white curve is an exact sinewave.

You might like to consider how the elliptical orbit's
curve will change with distance. For example can you
do the same at a distance where VDoppler and ADoppler
are of equat magnitude (the 45 degree case for a
circular orbit).


In recent measages you seem to have been switching
between saying VDoppler doesn't exist at all and saying
it exists but is negligible presumably because it is
much smaller than ADoppler. I have held off replying to
see if you would clarify that (and also I was out last
night and we had visitors at the weekend). I've also
been tinkering with a GUI and might do a simulation for
comparison with yours but I have a couple of other
projects I'm working on too so I may not spend too much
time duplicating what you've already done. Does your
program actually include VDoppler or not?

George

On 28 Mar 2007 02:16:59 -0700, "George Dishman"
wrote:

On 28 Mar, 08:10, HW@....(Henri Wilson) wrote:
On Sun, 25 Mar 2007 23:34:03 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message

George, when you can, have a look at

http://www.users.bigpond.com/hewn/ellip_circle.jpg

This shows how an elliptical orbit can produce a near perfect sine wave under
certain condition whilst the circular orbit produces nothing like one for
exactly the same parameter values.

Yaw angle is -90 (periastron closest to observer).

The white curve is an exact sinewave.

You might like to consider how the elliptical orbit's
curve will change with distance.

For small changes in magnitude I cannot tell the difference between the output
for a circular orbit and one with a small eccentricity.

The shape of the 'sinewave' produced for an elliptical orbit certainly changes
with distance, as expected. However it is certainly interseting to note that a
perfect sine wave 'bunching curve' can be produced by a star in an elliptical
orbit.

There is probably an algebraic reason for this ...but I don't think I'll bother
to find out what it is.

For example can you
do the same at a distance where VDoppler and ADoppler
are of equat magnitude (the 45 degree case for a
circular orbit).

George, I think what you are calling VDoppler is what you would get if you
placed a large number of equally spaced lights around a spinning wheel (Edge
on). Those on the sides would be 'VDoppler bunched' or separated.
This is not the situation we are examining. The pulses are emitted in sequence
and not all at the same instant..and not at exactly the same point.

In recent measages you seem to have been switching
between saying VDoppler doesn't exist at all and saying
it exists but is negligible presumably because it is
much smaller than ADoppler.

I have finally realised there is no VDoppler in the classical sense (as in the
case of the spinning wheel, above)

What the program measures is the rate at which pulses arrive. The ones on the
edge are emitted under constant velocity conditions and arrive at *very nearly*
the rate at which they are emitted. There is a very small difference due to the
fact that consecutive pulses are not emitted at the same point. I have
incorporated that by adding an Rsin(x) term to the star distance. It is
generally negligible.

I have held off replying to
see if you would clarify that (and also I was out last
night and we had visitors at the weekend). I've also
been tinkering with a GUI and might do a simulation for
comparison with yours but I have a couple of other
projects I'm working on too so I may not spend too much
time duplicating what you've already done. Does your
program actually include VDoppler or not?

George, I think your model is something like a spinning wheel with many lights
equally spaced around its rim. VDoppler shift will occur in that model, if you
assume constant light speed to the observer from all sources. The correct model
is a spinning wheel that has one *flashing* light on its rim. There is a subtle
difference. Conventional VDoppler does not occur in this case.
The shift in the former is (c+v)/c. In the latter it is something like
(D-Rsin(xt))/D and very soon disappears.

Do you see what I'm getting at?

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.

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:
On Sun, 25 Mar 2007 23:34:03 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message

George, when you can, have a look at

http://www.users.bigpond.com/hewn/ellip_circle.jpg

This shows how an elliptical orbit can produce a near perfect sine wave under
certain condition whilst the circular orbit produces nothing like one for
exactly the same parameter values.

Yaw angle is -90 (periastron closest to observer).

The white curve is an exact sinewave.

You might like to consider how the elliptical orbit's
curve will change with distance.

For small changes in magnitude I cannot tell the difference between the output
for a circular orbit and one with a small eccentricity.

What I expect is that as the distance changes, the ratio
of VDoppler to ADoppler changes giving a change in phase.
To compensate for that, you might need to alter the yaw
or a change of eccerntricity might do it.

The shape of the 'sinewave' produced for an elliptical orbit certainly changes
with distance, as expected. However it is certainly interseting to note that a
perfect sine wave 'bunching curve' can be produced by a star in an elliptical
orbit.

There is probably an algebraic reason for this ...but I don't think I'll bother
to find out what it is.

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.
theory

For example can you
do the same at a distance where VDoppler and ADoppler
are of equat magnitude (the 45 degree case for a
circular orbit).

George, I think what you are calling VDoppler is what you would get if you
placed a large number of equally spaced lights around a spinning wheel (Edge
on). Those on the sides would be 'VDoppler bunched' or separated.

I'm not sure I follow that but it is certainly not
what I am doing.

This is not the situation we are examining. The pulses are emitted in sequence
and not all at the same instant..and not at exactly the same point.
....
I have finally realised there is no VDoppler in the classical sense (as in the
case of the spinning wheel, above)

What the program measures is the rate at which pulses arrive. The ones on the
edge are emitted under constant velocity conditions and arrive at *very nearly*
the rate at which they are emitted. There is a very small difference due to the
fact that consecutive pulses are not emitted at the same point.

Right, that is the cause of classical VDoppler. Two
pulses emitted 2.295 ms apart travel slightly different
distances due to the motion of the source. At an orbital
speed of 27983 m/s when the pulsar is moving directly
towards us, the second pulse would travel 64.22 m less
than the first which corresponds to about 214 ns. The
VDoppler would be about 93 parts per million.

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 have held off replying to
see if you would clarify that (and also I was out last
night and we had visitors at the weekend). I've also
been tinkering with a GUI and might do a simulation for
comparison with yours but I have a couple of other
projects I'm working on too so I may not spend too much
time duplicating what you've already done. Does your
program actually include VDoppler or not?

George, I think your model is something like a spinning wheel with many lights
equally spaced around its rim.

No, it is what you describe above. You say you have
an R*sin(x) factor in the distance to address it,
though whether that works or not depends on your
code obviously.

VDoppler shift will occur in that model, if you
assume constant light speed to the observer from all sources. The correct model
is a spinning wheel that has one *flashing* light on its rim. There is a subtle
difference. Conventional VDoppler does not occur in this case.

You seem to have an odd idea of "conventional VDoppler",
the single flashing light on the rim of the wheel is how
I would think of it.

The shift in the former is (c+v)/c. In the latter it is something like
(D-Rsin(xt))/D and very soon disappears.

Do you see what I'm getting at?

Not really. The classical Doppler is c/(c-v) for a single source
that moves which is how the pulsar behaves. Balistic theory
changes the speed so it becomes (c+v)/c where v is the component
along the line of sight and includes the sin(theta) term.

I don't know where you get this idea of multiple sources.

George

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:
On Sun, 25 Mar 2007 23:34:03 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message

George, when you can, have a look at

http://www.users.bigpond.com/hewn/ellip_circle.jpg

This shows how an elliptical orbit can produce a near perfect sine wave under
certain condition whilst the circular orbit produces nothing like one for
exactly the same parameter values.
Yaw angle is -90 (periastron closest to observer).

The white curve is an exact sinewave.

You might like to consider how the elliptical orbit's
curve will change with distance.

For small changes in magnitude I cannot tell the difference between the output
for a circular orbit and one with a small eccentricity.

What I expect is that as the distance changes, the ratio
of VDoppler to ADoppler changes giving a change in phase.
To compensate for that, you might need to alter the yaw
or a change of eccerntricity might do it.

I assure you George, the VDoppler term is insignificant.

The shape of the 'sinewave' produced for an elliptical orbit certainly changes
with distance, as expected. However it is certainly interseting to note that a
perfect sine wave 'bunching curve' can be produced by a star in an elliptical
orbit.

There is probably an algebraic reason for this ...but I don't think I'll bother
to find out what it is.

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

Anyway, I have demonstrated my point...A circular orbit can produce a
non-sinelike brightness curve and an elliptical one can. J1909-3744 might have
a more eccentric orbit than claimed and PSR1913+16 has a nearly circular
orbit..or at least one with a far lower eccentricity..

For example can you
do the same at a distance where VDoppler and ADoppler
are of equat magnitude (the 45 degree case for a
circular orbit).

George, I think what you are calling VDoppler is what you would get if you
placed a large number of equally spaced lights around a spinning wheel (Edge
on). Those on the sides would be 'VDoppler bunched' or separated.

I'm not sure I follow that but it is certainly not
what I am doing.

This is not the situation we are examining. The pulses are emitted in sequence
and not all at the same instant..and not at exactly the same point.
...
I have finally realised there is no VDoppler in the classical sense (as in the
case of the spinning wheel, above)

What the program measures is the rate at which pulses arrive. The ones on the
edge are emitted under constant velocity conditions and arrive at *very nearly*
the rate at which they are emitted. There is a very small difference due to the
fact that consecutive pulses are not emitted at the same point.

Right, that is the cause of classical VDoppler. Two
pulses emitted 2.295 ms apart travel slightly different
distances due to the motion of the source. At an orbital
speed of 27983 m/s when the pulsar is moving directly
towards us, the second pulse would travel 64.22 m less
than the first which corresponds to about 214 ns. The
VDoppler would be about 93 parts per million.

.....but George, the velocity is NOT 27983 m/s.
It is more likely only a few metres/sec.

Your figure is that which would apply if all the ADoppler was assumed to be
VDoppler....and that's where astronomy has been getting it wrong for years....

....but that's beside the point for the present....

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.

It is treated differently in the constant c model and the BaTh.

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'm going to have to look at this in more detail George.
This is rather difficult stuff to program.

I have held off replying to
see if you would clarify that (and also I was out last
night and we had visitors at the weekend). I've also
been tinkering with a GUI and might do a simulation for
comparison with yours but I have a couple of other
projects I'm working on too so I may not spend too much
time duplicating what you've already done. Does your
program actually include VDoppler or not?

George, I think your model is something like a spinning wheel with many lights
equally spaced around its rim.

No, it is what you describe above. You say you have
an R*sin(x) factor in the distance to address it,
though whether that works or not depends on your
code obviously.

I've changed the starting point so that is R.cos(x)...
It comes out as almost negligible compared with the distance traveled.
Let me have a closer look at this. The problem is that the two effects are hard
to separate with the 'pulse bunch' method. I'm might be missing something
here....but cannot see what it is.

VDoppler shift will occur in that model, if you
assume constant light speed to the observer from all sources. The correct model
is a spinning wheel that has one *flashing* light on its rim. There is a subtle
difference. Conventional VDoppler does not occur in this case.

You seem to have an odd idea of "conventional VDoppler",
the single flashing light on the rim of the wheel is how
I would think of it.

OK.


The shift in the former is (c+v)/c. In the latter it is something like
(D-Rsin(xt))/D and very soon disappears.

Do you see what I'm getting at?

Not really. The classical Doppler is c/(c-v) for a single source
that moves which is how the pulsar behaves. Balistic theory
changes the speed so it becomes (c+v)/c where v is the component
along the line of sight and includes the sin(theta) term.

I don't know where you get this idea of multiple sources.

If the wheel was spinning very rapidly, the lights on the edges would appear
closer together on one side and further apart on the other.
CMIIW...

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.

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.



"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 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

On 29 Mar 2007 00:09:21 -0700, "George Dishman"
wrote:

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 distortion of the brightness curve for circular orbits looks quite
symmetrical. I tried varying the yaw angle very slightly but it skewed the
curve away from a sine wave.
I think the major axis has to be aligned witrh teh LOS. However, we don;t know
how acccurate the published curves are....so you are probably right.

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.

Well I can telll you one thing. The extinction distance is directly
proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.042 years..etc.
....always independent of peripheral velocity.

How can you explain THAT?


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.

The c+v factor might effectively 'squash' the ellipse back into a circle. That
seems plausible.

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.

Yes..and I agree there has to be a degree of extinction...even though the
distance anomaly can now be explained by the fact that the calculated velocity
curves might be grossly exaggerated.

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

I will upload my program to the website George so you can fiddle with it.
It is by no measns complete but is OK for circular orbits.

Click the red button after selecting eccentricity then click either the yellow
one (for my original method) or 'george' for your quick method.
George has the VDoppler correction included...
Increase 'output size' to see the curve at short distances.
If you hold the mouse button down, a vertical line appears on the screen to
compare phases.

http://www.users.bigpond.com/hewn/newvariables.exe

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 wrote:

The extinction distance is directly proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.42 years..etc.
...always independent of peripheral velocity.

How can you explain THAT?

As I said 19 and 20 March, the light speed unification
distance is inversely proportional to the rate of pulse
bunching. The more rapidly the pulses bunch, the shorter
the unification distance. What you have found is the
obvious fact that the rate of pulse bunching is inversely
proportional to the period. All else being equal, the
shorter the period, the more rapidly the pulses bunch.
So naturally, the shorter the period, the shorter the
unification distance.

Leonard

"Henri Wilson" HW@.... wrote in message
...
On 29 Mar 2007 00:09:21 -0700, "George Dishman"
wrote:
On 29 Mar, 01:32, HW@....(Henri Wilson) wrote:

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 distortion of the brightness curve for circular orbits looks quite
symmetrical. I tried varying the yaw angle very slightly but it skewed the
curve away from a sine wave.
I think the major axis has to be aligned witrh teh LOS. However, we don;t
know
how acccurate the published curves are....so you are probably right.

I've already replied but I didn't have time to explain
where the numbers came from so here's a bit more detail.

The residuals on the timing measurements are measured at
74ns compared with a pulse period of 2.295ms. They say
somewhere that if they can reconfigure the receivers to
make better use of multiple channels, they should be
able to get that down to 10ns.

If yaw distorts the shape rather than changing the phase,
a crude estimate of the phase accuracy is 74ns in 2.295ms
or about 3 parts per million, that is 0.011 degrees on
the phase.

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.

Well I can telll you one thing. The extinction distance is directly
proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.042 years..etc.
...always independent of peripheral velocity.

How can you explain THAT?

See my addition to Leonard's response, you require an
_incredible_ coincidence between the inclination from
which we are viewing the system and the properties of
space along the line of sight.

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

I will upload my program to the website George so you can fiddle with it.
It is by no measns complete but is OK for circular orbits.

Click the red button after selecting eccentricity then click either the
yellow
one (for my original method) or 'george' for your quick method.
George has the VDoppler correction included...
Increase 'output size' to see the curve at short distances.
If you hold the mouse button down, a vertical line appears on the screen
to
compare phases.

http://www.users.bigpond.com/hewn/newvariables.exe

I'll have a go over the weekend but it depends whether you
can set numbers sufficiently low. If extinction is around
one light minute as I suspect, the GUI is going to be
inconvenient.

George