From: David Evens on
On Thu, 16 Jun 2005 11:52:17 GMT, H@..(Henri Wilson) wrote:
>[Stop thinking rationaly about the consequences of my claims!]

What, and let you possibly delude the ignorant?
From: bz on
H@..(Henri Wilson) wrote in
news:u6q4b11bluqsrp2phed5l457df9ie0oc4g(a)4ax.com:

> On Thu, 16 Jun 2005 22:32:06 +0000 (UTC), bz
> <bz+sp(a)ch100-5.chem.lsu.edu> wrote:
>
>>H@..(Henri Wilson) wrote in
>>news:hco3b19rlge9s4ss0nhatfn5lpj75rbfjc(a)4ax.com:
>>
>
>>
>>>>The problem is that for any practicle size for the orgiting star(s),
>>>>the orbital velocity will be MUCH higher than either 21 or 17 km/s to
>>>>orbit in 5.36 days (much less the 3.7 days that RT Aur shows).
>>>
>>> have you considered that the orbit plane might be nearly perpendicular
>>> to the LOS.....no!
>>
>>That doesn't help the stars orbit faster.
>>Try putting the correct orbital velocity in AND the KNOWN orbital roll
>>values.
>
> Take an orbit with a circumference of 300 million kms.
>
> To orbit once per day, the star would have to travel at 3E8/3E5/86400
> which is about 0.012c
>
> To orbit every five days, 0.0023c
>
> Fast but very possible.
>
> You woudn't want to get in the way of one!


I don't find a self consistent set of answers for these values.
What masses are you using, eccentricity, major & minor axis, etc.

(I get strange results, like a & b each > the perimeter, which is, of
course, impossible. The radius can't be greater than the circumference.
This probably means that some values are going imaginary on me.)


>>> Radial velocity is the component in the observer direction. It is not
>>> the peripheral velocity.
>>
>>There are two points on the orbit where radial velocity is equal to the
>>component in the observers direction. The point of max velocity toward
>>and max velocity away from are the points where the tangent to the
>>eliplse points directly toward or away from the observer.
>>
>>At all other points, the radial velocity will be lower.
>
> Paul Andersen once defined the 'radial velocity' as the peripheral
> velocity component in the direction of the observer.
> Doppler shift measurements indicate radial velocity directly.

That is correct. And at two points in the orbit, the radial velocity will
be equal to the orbital velocity, as I described above.

>>>>>>The larger the mass, the greater the orbit's diameter AND the
>>>>>>greater the orbital velocity. When M2 is 3e9 times M1, the orbital
>>>>>>velocity reaches c. With an orbital diameter of 1.5e10 km.
>>>>>
>>>>> You have something wrong there.
>>
>>>>Why do you say that?
>>
>>
>>>>You ran the numbers and you came out with different numbers?
>>>>You just don't like my numbers?
>>>>Perhaps you don't believe the formula for orbital period?
>>>>P=2 pi sqrt(a^3/(G*(M1+M2))
>>
>>> I do.
>>> I think it is only correct for M1>> M2.
>>
>>Look at the formula again.
>>
>>The formula for M1>>M2 neglects to add the value of M2 because it won't
>>make a difference.
>>
>>The formula I gave is for ANY values of M1 and M2.
>
> The formula you gave it the same as the one for M1>>M2, in which case
> the larger star remains at one focus and the smaller one rotates in
> elliptical orbit.

There is no assumption as to either star remaining at the barycentre.

The TOTAL MASS appears to be, effectively, at the barycentre.

> I don't see quite why the same applies when the barycentre constantly
> moves by a large amount,

The barycentre does NOT move. You see the stars motions wrt the barycentre.

> as it does when the stars are both about the
> same size.

The earth/moon barycentre follows an eliptical orbit around the sun.
The earth and moon orbit that barycentre so the earths orbit around the sun
'wiggles' every 28 days.

> I'm not saying it is wrong, (Kepler didn't make many
> mistakes) I just want to fully understand how the shape of the orbit is
> affected. Maybe the velocity increase at the perihelion exactly
> compensates for the slowing at the aphelion.

Perhaps you understand a bit better now?

> My program uses true elliptical orbits without asking any questions as
> to why there are as they are..

and without any checking to see if all the values are consistent with each
other. Making for some very strange looking light curves, at times.

>>> This is worth checking.
>>> Why don't you write a little program to plot elliptical orbits and see
>>> if it agrees.
>>
>>I don't doubt the formula.
>
> fair enough.
>
>
>>>>That depends on ones model. I am afraid that even you will need to
>>>>admit that it is difficult to get one star to orbit another at an
>>>>orbital velocity > c.
>>>
>>> I don't see why.
>>
>>Look at the orbital parameters you would need to approach c.
>
> See above.
>
>>
>>And, you have never argued that a particle with rest mass can reach c.
>>Have you changed your idea?
>>
>>You are not going to claim that the 'Henri Reverse field' bubble will go
>>away just because we are dealing with something that has the mass of a
>>star, are you?
>
> No. not with stars. One was probably captured by the other. or one large
> one broke into two pieces.
> Incidentally, my program 'threebody.exe' shows how captures occur. They
> require a third body, otherwise the captured one either flies off again
> or ends up colliding with the other. So it is quite feasible that a star
> could end up rotating around another with a velocity <c.

as long as you say <c, I agree.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: Henri Wilson on
On Thu, 16 Jun 2005 22:32:06 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu>
wrote:

>H@..(Henri Wilson) wrote in
>news:hco3b19rlge9s4ss0nhatfn5lpj75rbfjc(a)4ax.com:
>

>.....
>
>>>The problem is that for any practicle size for the orgiting star(s), the
>>>orbital velocity will be MUCH higher than either 21 or 17 km/s to orbit
>>>in 5.36 days (much less the 3.7 days that RT Aur shows).
>>
>> have you considered that the orbit plane might be nearly perpendicular
>> to the LOS.....no!
>
>That doesn't help the stars orbit faster.
>Try putting the correct orbital velocity in AND the KNOWN orbital roll
>values.

Bob, no telescope can resolve any of these binaries. They resemble small dots,
even on HST. The only info we have is their radial velocities from doppler
shift. The 'roll' is not known. (deviation from 'edge-on')


>
>> Radial velocity is the component in the observer direction. It is not
>> the peripheral velocity.
>
>There are two points on the orbit where radial velocity is equal to the
>component in the observers direction. The point of max velocity toward and
>max velocity away from are the points where the tangent to the eliplse points
>directly toward or away from the observer.

That is correct.
These are where the maximum c+v and minimum c-v occur. For any particular
ellipse, the values depend on the yaw angle.
'Roll' just applies a cosine factor to thes values.

>
>At all other points, the radial velocity will be lower.

NO!!!
It's lowest value is c-v.

There are also two pooints where the radial velocity is zero towards the
observer. The lightspeed is then c.
You can see this in the 'lightfront' section of my program. Two points on the
curves remain at the same relative positions. (allow for the initial slope)


>
>
>>>You ran the numbers and you came out with different numbers?
>>>You just don't like my numbers?
>>>Perhaps you don't believe the formula for orbital period?
>>>P=2 pi sqrt(a^3/(G*(M1+M2))
>
>> I do.
>> I think it is only correct for M1>> M2.
>
>Look at the formula again.
>
>The formula for M1>>M2 neglects to add the value of M2 because it won't make
>a difference.
>
>The formula I gave is for ANY values of M1 and M2.
>
>> This is worth checking.
>> Why don't you write a little program to plot elliptical orbits and see
>> if it agrees.
>
>I don't doubt the formula.
>
>....
>
>>>That depends on ones model. I am afraid that even you will need to admit
>>>that it is difficult to get one star to orbit another at an orbital
>>>velocity > c.
>>
>> I don't see why.
>
>Look at the orbital parameters you would need to approach c.
>
>And, you have never argued that a particle with rest mass can reach c.
>Have you changed your idea?
>
>You are not going to claim that the 'Henri Reverse field' bubble will go away
>just because we are dealing with something that has the mass of a star, are
>you?


HW.
www.users.bigpond.com/hewn/index.htm

Sometimes I feel like a complete failure.
The most useful thing I have ever done is prove Einstein wrong.
From: Henri Wilson on
On Fri, 17 Jun 2005 12:51:25 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu>
wrote:

>H@..(Henri Wilson) wrote in
>news:u6q4b11bluqsrp2phed5l457df9ie0oc4g(a)4ax.com:
>

>>>
>>>That doesn't help the stars orbit faster.
>>>Try putting the correct orbital velocity in AND the KNOWN orbital roll
>>>values.
>>
>> Take an orbit with a circumference of 300 million kms.
>>
>> To orbit once per day, the star would have to travel at 3E8/3E5/86400
>> which is about 0.012c
>>
>> To orbit every five days, 0.0023c
>>
>> Fast but very possible.
>>
>> You woudn't want to get in the way of one!
>
>
>I don't find a self consistent set of answers for these values.
>What masses are you using, eccentricity, major & minor axis, etc.

I'm assuming the central mass is >> than the orbiting one and that the orbit is
roughly circular. In that case, the orbit radius is almost independent of mass.

Our sun has a radius of 0.7E6 kms and is around 150 million kms away.
The surface of a star 40 times larger than our sun and moving in a circular
orbit of circumference 314 million kms would still be about 22 million kms from
the centre of the circle.

>
>(I get strange results, like a & b each > the perimeter, which is, of
>course, impossible. The radius can't be greater than the circumference.
>This probably means that some values are going imaginary on me.)

Check your figures.
Use a circumference of 314 million kms.


>> Paul Andersen once defined the 'radial velocity' as the peripheral
>> velocity component in the direction of the observer.
>> Doppler shift measurements indicate radial velocity directly.
>
>That is correct. And at two points in the orbit, the radial velocity will
>be equal to the orbital velocity, as I described above.

and two points will have zero radial velocity towards the observer.
So four point around any orbit will have radial velocities c, c+v, c and c-v.
The cosine correction for 'roll' only operates on the v part.

>
>>>>>>>The larger the mass, the greater the orbit's diameter AND the
>>>>>>>greater the orbital velocity. When M2 is 3e9 times M1, the orbital
>>>>>>>velocity reaches c. With an orbital diameter of 1.5e10 km.
>>>>>>
>>>>>> You have something wrong there.
>>>
>>>>>Why do you say that?
>>>
>>>
>>>>>You ran the numbers and you came out with different numbers?
>>>>>You just don't like my numbers?
>>>>>Perhaps you don't believe the formula for orbital period?
>>>>>P=2 pi sqrt(a^3/(G*(M1+M2))
>>>
>>>> I do.
>>>> I think it is only correct for M1>> M2.
>>>
>>>Look at the formula again.
>>>
>>>The formula for M1>>M2 neglects to add the value of M2 because it won't
>>>make a difference.
>>>
>>>The formula I gave is for ANY values of M1 and M2.
>>
>> The formula you gave it the same as the one for M1>>M2, in which case
>> the larger star remains at one focus and the smaller one rotates in
>> elliptical orbit.
>
>There is no assumption as to either star remaining at the barycentre.
>
>The TOTAL MASS appears to be, effectively, at the barycentre.
>
>> I don't see quite why the same applies when the barycentre constantly
>> moves by a large amount,
>
>The barycentre does NOT move. You see the stars motions wrt the barycentre.

I didn't actually doubt it. I just want to understand more clearly why, for a
particular orbit shape, stars of similar size have twice the period of a very
small object orbiting a much larger one.

>
>> as it does when the stars are both about the
>> same size.
>
>The earth/moon barycentre follows an eliptical orbit around the sun.
>The earth and moon orbit that barycentre so the earths orbit around the sun
>'wiggles' every 28 days.

Yes, the barycentre follows a cycloidal path.

>
>> I'm not saying it is wrong, (Kepler didn't make many
>> mistakes) I just want to fully understand how the shape of the orbit is
>> affected. Maybe the velocity increase at the perihelion exactly
>> compensates for the slowing at the aphelion.
>
>Perhaps you understand a bit better now?

I want to plot how orbit shape and velocity changes as the mass of M1 is
increased from <<M2 to =M2.

>
>> My program uses true elliptical orbits without asking any questions as
>> to why there are as they are..
>
>and without any checking to see if all the values are consistent with each
>other. Making for some very strange looking light curves, at times.

No. We know that the two members of a binary pair are in elliptical orbits of
the same eccentricity. There phase is 180 out. The ratio of major axes is
proportion to M1/M2. (CMIIW)


>>>You are not going to claim that the 'Henri Reverse field' bubble will go
>>>away just because we are dealing with something that has the mass of a
>>>star, are you?
>>
>> No. not with stars. One was probably captured by the other. or one large
>> one broke into two pieces.
>> Incidentally, my program 'threebody.exe' shows how captures occur. They
>> require a third body, otherwise the captured one either flies off again
>> or ends up colliding with the other. So it is quite feasible that a star
>> could end up rotating around another with a velocity <c.
>
>as long as you say <c, I agree.

:~)
Genuine typo!!!!!!!!!
I meant >c


HW.
www.users.bigpond.com/hewn/index.htm

Sometimes I feel like a complete failure.
The most useful thing I have ever done is prove Einstein wrong.
From: Henri Wilson on
On Thu, 16 Jun 2005 22:18:43 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu>
wrote:

>H@..(Henri Wilson) wrote in
>news:klp3b11khrm3bhq2ajkfef1k0991ncnktf(a)4ax.com:
>

>>>{which would mean that you are not really seeing an extinction distance}
>>>I am not sure that your program has any factors in it associated with
>>>extinction. Therefore I am very suspicious of your conclusion.
>>
>> Why.
>
>Because you don't seem to have anything built into your program that can
>give you 'an extinction distance' while you do seem to have many problems
>built into your program, any one of which may produce strange results
>because you don't have any 'sanity checks' built into the interdependent
>values.

Bob, my program does not produce anything unusual.
The 'Lightfront' section show exactly how hypothetical light pulses emitted at
equal intervals around a star's orbit will move relatively as they move away at
c+v, where v is the star's radial velocity component towards the observer at
the instant of emission..

The brightness curves that are generated can be seen to match the lightfon
curves exactly. A brightness peak will occur at points where the lightfront
curve is vertical. You can see how many multiple images are expected at a
particular distance.

......but now I have good evidence that extinction kicks in and unifies the
speed of all the light leaving the star. That might occur within about ten LYs.

>> The evidence is too striking to be mere coincidence.
>
>No. Throughout the history of science there have been MANY 'coincidences'.
>I can tell you from personal experience that it is very easy to missread
>evidence and get all excited over a 'new discovery' only to find out that
>one has either made a mistake or has rediscovered an old discovery.

Bob, doesn't the fact that so many light curves can be matched by the BaT make
you wonder even slightly?

Contrary to what the desperate Andersen claims, my program cannot produce any
curve I want. It is based solely on how slow slight emitted from a precise
elliptical orbit will be overtaken by faster light. The maths are mainly double
precision. It is a very accurate representation of what should happen.

I have emphasized that the shapes of the predicted brightness curves are the
revealing feature.. ..not the actual parameter values.

>
>> I have always had in my mind the problem of extinction as light tavels
>> through space. That was, after all, the basis of the refutation of
>> DeSitter's arguments against the BaT.
>> Now I have the evidence.
>
>Calm down and look everything over very carefully before you go jumping to
>such conclusions.

Well, Bob, I'll ask you an intelligent question and give you one to ask ME in
return.

1) How is it that the brightness curves of eclipsing binaries are not flat away
from the 'eclipse dip'.

2) if extinction unifies light speed after a certain distance, (as Henri
claims) how can the observed doppler shift still reveal different radial
velocities? Similarly, if light arrives outside the Earth's atmosphere at
different speeds but all these speeds are unified to c before it reaches the
ground, how should doppler shift be affected?


>
>....
>
>>>Anything which removes some of the problems from the program is good.
>
>> It doesn't really have problems. ...complications maybe...Yaw angle is
>> one.
>
>There are several problems associated with that 'complication'.

It is becoming simpler. I fixed the Pause/continue button.



HW.
www.users.bigpond.com/hewn/index.htm

Sometimes I feel like a complete failure.
The most useful thing I have ever done is prove Einstein wrong.