Mercury's Perihelion Precession

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />For the sun moving around the barycenter of the solar system, the coordinate system fixed with the sun does of course not rotate but is nevertheless accelerated as it does not move with a constant velocity vector. This acceleration has to be accounted for when changing the system of reference from heliocentric to barycentric or vice versa (as indicated in my mathematical analysis further above).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You are mistaken. As long as the motions are determined in an inertial coordinate system (which they are), you can then switch the origin of coordinates to a fly zipping around at random without any new consequences for the orbits of bodies.

Rotating coordinates would create fictitious, apparent Coreolis and/or centrifugal forces for the reason I described. But we do not have rotating coordinates here. We simply have a conversion from a fixed origin to a translating origin with an irregular path. Switching origins without a rotation does not introduce pseudo-forces because the dynamics were all determined in an inertial reference frame before the origin switch.

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Principia:

HYPOTHESIS I.
That the center of the system of the world is immovable.
This is acknowledged by all, while some contend that the earth, others that the sun is fixed in that center. Let us see what may from hence follow.


PROPOSITION XI. THEOREM XI.
That the common center of gravity of the earth, the sun, and all the planets, is immovable.

For (by Cor. 4 of the laws) that center either is at rest, or moves uniformly forward in a right line; but if that center moved, the center of the world would move also, against the Hypothesis (I).


Looking for the center of the world people?

LOL. You need relativistic dynamics.

Makis

Makis, How do you derive the precession of Mercury's orbit using GR? I'm still stuck on this detail because I don't know how to determine the cause of the 575"/yr total. Do you know the math behind this calculation? And then how is the 1575 meter radius of the sun used in getting the remaining precession?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">what you need in order to define a change of the longitude of Mercury's perihelion is a description of the orbit in terms of the classical orbital elements, which obviously is based on the assumption that you are dealing with ellipses. However, ellipses only exist for two-body problems but not for 3 or more bodies. So trying to fit the data in terms in ellipses (even if corrections are applied) will inevitably lead to deviations from the actual orbit if you are just looking at effects small enough.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This is all incorrect. The perihelion point (point of closest approach to the Sun, by definition) is well-defined and, by itself, defines the major apsis that advances. So perihelion motion is well-defined regardless of the type of orbit or complexity of the motion.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The longitude of the perihelion is only one of six parameters to characterise an (elliptical) orbit (see www.astro.uu.nl/~strous/AA/en/reken/hemelpositie.html ) and as you mentioned yourself earlier in this thread, these quantities are obtained by the best fit to the observations (and this fit rests obviously on the assumption that one is dealing with an ellipse). In reality however the orbit is not elliptical due to the various disturbances involved and the perihelion is therefore by no means well defined as the point of closest approach to the sun (in principles there could be even two perihelions if a disturbance makes a corresponding 'dent' in the orbit).
The point is that any apparent increase of speed of a planet in its orbit would have to be interpreted as a precession effect if the orbit is interpreted in terms of an ellipse for which the orbital parameters are known and fixed. As shown in my mathematical analysis earlier in this thread however, such an apparent increase of orbital speed can be simply produced by the transformation from the heliocentric to the barycentric system and may not have anything to do at all with a precession. If you look at Morrison and Ward's paper, the 'perihelion precession' is indeed merely inferred from the time discrepancies for a number of Mercury-Sun transits and does not actually compare perihelion positions at all.

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<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />The longitude of the perihelion is only one of six parameters to characterise an (elliptical) orbit ... and as you mentioned yourself earlier in this thread, these quantities are obtained by the best fit to the observations (and this fit rests obviously on the assumption that one is dealing with an ellipse). In reality however the orbit is not elliptical due to the various disturbances involved and the perihelion is therefore by no means well defined as the point of closest approach to the sun ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I continue to be amazed that you think the last 200 years of celestial mechanics experts were naive, and that only your genius has finally set things straight. Surely you have some sense of how unlikely that is to be reality?

First, your whole scenario here is irrelevant because, in the two-body problem of a point particle in orbit around a source mass, the orbit is exactly elliptical. Yet if the particle lies in Mercury's orbit and the source mass is the Sun, there are no planetary perturbations so the 532"/cy precession (and like changes in other elements) go away, but the 43"/cy GR precession of the perfect ellipse still exists and is still well-defined.

Second, the first thing one learns after studying the 2-body problem is that, when there are perturbing planets, we define an "osculating ellipse": the true elliptical orbit at some instant of time that the planet would follow if all perturbations were suddenly removed. This osculating ellipse describes the position and motion very well near that instant. And at later times, one need only calculate the small, slow changes in the osculating ellipse caused by the planets and other perturbations. One then has a succession of true ellipses that represent the actual orbit. And one can choose either "osculating elements" to have the exact ellipse at any desired instant, or "mean elements" to track the average behavior of the planet.

Any way you look at it, nobody knowledgable is leaving out anything important to the dynamics. The calculated orbits are as exact as the number of decimal places we choose to carry will allow. -|Tom|-

The orbit of Mercury is generated at JPL/horizons and it seems the precession of perihelion is several degrees per century. I was told a degree was 3600" so how can this be true? Am I doing something else wrong? There is nothing about the cause of this posted at JPL.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
First, your whole scenario here is irrelevant because, in the two-body problem of a point particle in orbit around a source mass, the orbit is exactly elliptical. Yet if the particle lies in Mercury's orbit and the source mass is the Sun, there are no planetary perturbations so the 532"/cy precession (and like changes in other elements) go away, but the 43"/cy GR precession of the perfect ellipse still exists and is still well-defined.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">In GR the 43" precession still exists in this case but not in reality (all orbits in a 1/r^2 force field have to be closed (see Landau and Lifshi(t)z, Mechanics)).


<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Second, the first thing one learns after studying the 2-body problem is that, when there are perturbing planets, we define an "osculating ellipse": the true elliptical orbit at some instant of time that the planet would follow if all perturbations were suddenly removed. This osculating ellipse describes the position and motion very well near that instant. And at later times, one need only calculate the small, slow changes in the osculating ellipse caused by the planets and other perturbations.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">All right, but the problem is to calculate the perturbations exactly. Any approximation will introduce a certain error into the orbit calculation and I would guess that the approximations used are as usual of first order or at best of second order here (see for instance ads.harvard.edu/books/1989fcm..book/Chapter9.pdf and scienceworld.wolfram.com/physics/OrbitalPerturbation.html ).


<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The calculated orbits are as exact as the number of decimal places we choose to carry will allow<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Can you give me a reference for this? As far as I can see, the calculated orbit can only be as exact as the highest order of the perturbation theory allows. If a second order approximation is used, using a greater numerical accuracy would thus make the solution converge towards this 'second order' orbit but not towards the true orbit.

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