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Natural Science Forum / Physics / General Physics / April 2007Precession of Binary Star PSR 1913+16
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 in http://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) The combined orbit circumference is: C = 2 * (2*pi*r) The orbit time is: t = C / v The precession rate in orbits per orbit, where e is the average orbit eccentricity (0.617131): n = 2*pi*(1 - cos(asin(v/c))) / (1 - e*e) Converted to degrees per Earth year: d = n * (365/(t/24/60/60)) * 360 = 4.4, where modern General Relativity provides an approximation of 4.2 degrees per year. If you are interested, I have put up a short PDF document online, which includes C++ source code: http://cavekitty.com/archives/The_Relativistic_Precession_of_Binary_Star_PSR_191 3+16.zip Thank you for any information that you can provide. - Shawn >I am wondering if the following method for calculating the > relativistic precession of a highly eccentric orbit has a name, or is [quoted text clipped - 9 lines] > 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 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). 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. 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. 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. - Shawn On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> wrote: > <shala...@gmail.com> wrote in messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com... > >I am wondering if the following method for calculating the [quoted text clipped - 26 lines] > Wrong. What is the circumference of an ellipse? > http://mathworld.wolfram.com/images/gifs/EllipticGears.gif > Hi Androcles, > [quoted text clipped - 11 lines] > 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 > [quoted text clipped - 30 lines] >> Wrong. What is the circumference of an ellipse? >> http://mathworld.wolfram.com/images/gifs/EllipticGears.gif Like I said, I am only guessing because I don't have the data. Call it a prediction if you like. As for being a scientist, I'm fine with not fitting the bill. The methods I'm using are on par with General Relativity, regardless of what I'm called. In fact, all of the scientists that I've tried to communicate with have obviously lost the flare that comes with learning and exploring the world. Otherwise, I would assume that they would be interested in helping others learn about the universe instead of berating or ignoring them. I'm sure you can agree that it's lugubrious to want to associate oneself with that type of person, regardless of title. - Shawn On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> wrote: > <shala...@gmail.com> wrote in messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com... > > Hi Androcles, [quoted text clipped - 90 lines] > >> Wrong. What is the circumference of an ellipse? > >> http://mathworld.wolfram.com/images/gifs/EllipticGears.gif I should be more specific here. I am only commenting on the behaviour of those scientists at the university which I have paid good money toward. I wouldn't have wasted my time and effort if I had known they were going to be so completely useless to me in my desire to learn. I know it's final examinations right now, but taking a month to reply to an email message is stretching the limits of my good patience -- and I'm a parent, so please believe me when I say that I have a near- infinite supply of patience when it's appropriate. - Shawn On Apr 28, 7:57 pm, shala...@gmail.com wrote: > Like I said, I am only guessing because I don't have the data. Call it > a prediction if you like. [quoted text clipped - 108 lines] > > >> Wrong. What is the circumference of an ellipse? > > >> http://mathworld.wolfram.com/images/gifs/EllipticGears.gif >I should be more specific here. I am only commenting on the behaviour > of those scientists at the university which I have paid good money [quoted text clipped - 4 lines] > I'm a parent, so please believe me when I say that I have a near- > infinite supply of patience when it's appropriate. Under discussion is "Precession of Binary Star PSR 1913+16", not exam time, parenting, cost of tuition or tardiness of faculty. I'm not interested in your personal problems of everyday life no matter how specific you are or what excuses you make, to me they are mere gossip, and nor can I render you any assistance in resolving them. Show that the advance of longitude of perihelion of Mercury is not completely satisfied by Newtonian Mechanics. Show that PSR 1913+16 is binary. > On Apr 28, 7:57 pm, shala...@gmail.com wrote: >> Like I said, I am only guessing because I don't have the data. Call it [quoted text clipped - 109 lines] >> > >> Wrong. What is the circumference of an ellipse? >> > >> http://mathworld.wolfram.com/images/gifs/EllipticGears.gif On Apr 28, 5:57 pm, shala...@gmail.com wrote: [...] Don't listen to Androcles. He is a spewing idiot with delusions of competence. > Like I said, I am only guessing because I don't have the data. Call it > a prediction if you like. [quoted text clipped - 7 lines] > lugubrious to want to associate oneself with that type of person, > regardless of title. I am interested in helping homo sapiens sapiens to think about Nature as it is, I am not interested in training homo neanderthalensis to become a shaman pushing buttons on a calculator, dividing by zero and claiming he's now a mathematician. GR is on a par with palmistry, reading horoscopes and tea leaves. Learning about Nature does not mean inventing it. You choose to be a shaman, attempting to fool others as you succeed in fooling yourself. Whether you agree or not, it's idiotic to want to associate oneself with that type of imbecile, regardless of title. Better to ignore them. Please go away, start a new newsgroup, "shaman.physics", you are fine by not fitting the bill, and you don't. > On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> > wrote: [quoted text clipped - 92 lines] >> >> Wrong. What is the circumference of an ellipse? >> >> http://mathworld.wolfram.com/images/gifs/EllipticGears.gif P.S. Thank you again for the links. My main desire at the moment is for actual system data like eccentricity, apastron/periastron separation distances, and precession estimates/observations. However, I do find the pictures of these systems on NASA's page to be fascinating. On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> wrote: > <shala...@gmail.com> wrote in messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com... > > Hi Androcles, [quoted text clipped - 90 lines] > >> Wrong. What is the circumference of an ellipse? > >> http://mathworld.wolfram.com/images/gifs/EllipticGears.gif > P.S. Thank you again for the links. My main desire at the moment is > for actual system data like eccentricity, apastron/periastron > separation distances, and precession estimates/observations. Totally impossible to obtain. Anything you think you have is idle guesswork. > However, > I do find the pictures of these systems on NASA's page to be > fascinating. Where your heart line reaches and crosses your life line there'll be a crux in your life, you'll be at a crossroads. http://www.handanalysis.co.uk/ How fascinating...<yawn> "That fellow seems to me to possess but one idea, and that is a wrong one."--Dr. Samuel Johnson "There are well-dressed foolish ideas, just as there are well-dressed fools."--Nicolas Chamfort > On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> > wrote: [quoted text clipped - 92 lines] >> >> Wrong. What is the circumference of an ellipse? >> >> http://mathworld.wolfram.com/images/gifs/EllipticGears.gif The precession data is not impossible to obtain. I'm sure there's been one or two binaries found since 1973. As for my "problems", they are not personal, they're professional. I just got a very detailed reply from my university, which means that they were not ignoring me. I guess that means the problem is theirs, since they are overworked. By the way, where did you get the precession data that you used to verify your methods? Or have you only explained the precession of Mercury? - Shawn > The precession data is not impossible to obtain. I'm sure there's been > one or two binaries found since 1973. Then do it. Do what? Contact the local universities to see if I can get data on binary stars? I've already done that. How do you not get dizzy from the circles you're constantly running in? Don't get me wrong, I enjoy reading your posts. However, I'm confident at this point that you are constructing a fantasy conversation in your head that has nothing to do with what we're actually talking about. Not exactly frustrating, just bewildering. - Shawn On Apr 30, 1:22 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> wrote: > <shala...@gmail.com> wrote in messagenews:1177958128.065520.238070@y80g2000hsf.googlegroups.com... > > The precession data is not impossible to obtain. I'm sure there's been > > one or two binaries found since 1973. > > Then do it. > Do what? Contact the local universities to see if I can get data on > binary stars? I've already done that. It was you that said it was possible to obtain the data, not I. Run around in circles until you have it. > How do you not get dizzy from the circles you're constantly running > in? It's your circle. I'm sitting back waiting for you to obtain some data. <shrug> > Don't get me wrong, I enjoy reading your posts. However, I'm confident > at this point that you are constructing a fantasy conversation in your > head that has nothing to do with what we're actually talking about. > Not exactly frustrating, just bewildering. > - Shawn Are you? > On Apr 30, 1:22 pm, "Androcles" <Engin...@hogwarts.physics.co.uk> > wrote: [quoted text clipped - 3 lines] >> >> Then do it. |
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