Re: Precession of Binary Star PSR 1913+16


<shalayka@xxxxxxxxx> wrote in message news:1177792048.598466.293320@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Hi Androcles,

If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).

However, to perform my due diligence, and appease your objections, I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it exclusively
in my other endeavours.

What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would simplify
back down to a Sun-Mercury type relationship).


Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.



I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the equation. I
just can't seem to find a lot of information on binaries at this time.

Guessing is what horse race punters do. The bookie sets the odds and
wins every time. Mathematicians do not guess, they prove.


I appreciate your response, because the points you bring up are very
important to note, ex: eccentricity has a major part to play in this
equation.

No circular orbit can show precession.


It might be a little pugnacious of me to say so, but I'm not going to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.

You can have my time but it goes with my skepticism and incredulity.
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.
I am not willing to accept the forerunner of all binaries is a binary, based
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm

If you wish to call a star and planet a binary, then Sirius is a star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a bit
futher away than Saturn (29 years) and closer than Uranus (84 years).

You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of
pseudo-scientists.


- Shawn






On Apr 28, 12:09 pm, "Androcles" <Engin...@xxxxxxxxxxxxxxxxxxxxxx>
wrote:
<shala...@xxxxxxxxx> wrote in messagenews:1177782625.497884.294940@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or is
available in some published document?

Using the system parameters found inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:

The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r = (1576800e3 + 373300e3) / 2.

The average orbit velocity, where M is an average mass of 1.414 solar
masses:
v = sqrt(G*M / r)

That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the top.
How fast must it descend the same distance the other side to have an
average velocity of 60 mph?

I drive from home to London (30 miles for 1 hour), at 80 mph most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?

The combined orbit circumference is:
C = 2 * (2*pi*r)

Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif


.

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