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I don't think there's a way to meaningfully connect your comment to general relativity (GR). "The answer" in GR is that Mercury is in geodesic motion ("free fa
by raattgift 3y ago
I don't think there's a way to meaningfully connect your comment to general relativity (GR). "The answer" in GR is that Mercury is in geodesic motion ("free fall"), with its geodesic picked out by the mass of the solar system and principally its central mass (the sun). For all practical purposes, nothing about Mercury itself selects which geodesic its on.
The weak equivalence principle (WEP) says that in curved spacetime a small freely falling object's orbit around a large central pointlike mass is completely determined by the former's initial position in spacetime and its initial velocity.
The strong equivalence principle (SEP) says that this remains true for the smaller body even if that body is bound by its own self-gravitation: the orbit is determined fully by the initial position and velocity and not by the small body's internal composition.
We can of course replace the large-mass pointlike generator of a (exterior) Schwarzschild-like spacetime with an extended body that generates some perturbation of a more general central-mass spacetime (like Kerr or Kerr-Newman).
The WEP has been tested extensively in terrestrial labs (torsion balances) and satellites in Earth orbit (e.g. MICROSCOPE).
The SEP has been tested extensively using satellite and lunar laser ranging, and is supported by astrophysical observations including the triple-relativistic-star system comprising an inner white dwarf and PSR J0337+1715 orbited by an outer white dwarf.
That the SEP holds up so well means that we can replace Mercury with any mass much smaller than the sun (it does not have to be as small or smaller than Mercury) at any density and with any internal configuration as long as it is self-bound gravitationally and/or electromagnetically. So we could replace Mercury with a small black hole or a "hot Jupiter" and the orbit would be identical.
The sun generates a perturbed Kerr metric that must take into account not just its rotation but the sun's non-sphericity (the "solar bulge" makes it slightly oblate). Given the metric (or a good enough approximation) we can solve the geodesic equation, and see that Mercury's orbit follows one of the generated geodesics to high precision.
In <http://dx.doi.org/10.1103/PhysRevLett.120.191101 http://dx.doi.org/10.1103/PhysRevLett.120.191101> Clifford Will writes: "Finally, at a purely pedagogical level, it is often stated that the relativistic perihelion advance of Mercury is really only a test of the vacuum Schwarzschild solution (or of the slow rotation limit of the vacuum Kerr solution, if one wishes to include the frame-dragging effect), since all the relativistic effects can be derived simply from those metrics." Note that Will's paper uses post-Newtonian corrections to the 2PN level, which is perfectly reasonable given how small Mercury's v/c^2 is, and the ease with which the approach deals with perturbations from the other planets. See also Will's 1986 book, chapter 5 of which is devoted to Mercury's orbit.
> Mercury is heavier when it is at perihelion ... because it's moving faster
"Moving faster" is not a frame-independent statement. We can always use a freely falling coordinate system where Mercury is always at the origin, or a freely-falling coordinate system where in the neighbourhood of a point Mercury experiences no acceleration against those coordinates. There are of course an infinite number of systems of coordinates in which Mercury, the Sun, or both accelerate(s) against that set of coordinates over the course of an Earth year. The point of relativity is that physics do not depend on a choice of coordinates.
"Heavier" is at best ambiguous, requires a lot of care in stating it covariantly (compare the stress-energy tensor), and is in any event irrelevant if the Strong Equivalence Principle holds. More technically, the backreaction of Mercury on the metric generated by the sun (or sun + other planets) is negligible.
If you explain what you mean by Mercury's "relativistic mass" and roughly the magnitude you think its change should be through Mercury's orbit, someone might be help clear up what is probably a misconception. Bear in mind that Mercury moves very slowly compared to c.
Finally, with respect to your last paragraph:
Mercury orbital eccentricity: 0.205630. Pluto orbital eccentricity: 0.2488. If Earth has no perihelion precession because of the low eccentricity (0.01671) of Eearth's orbit, and Mercury's perihelion precession is driven by its higher eccentricity, what do you think Pluto's perihelion precession should be: higher or lower than Mercury's? And why?
- raattgift 3y agoETA: huh, sorry, this seems to have turned into something more like a blog entry than a brief self-correction. Obvious editing errors in the last paragraph - fixes italicized: "If Earth has an order of magnitude smaller perihelion precession ... Earth's orbit". The first fix was meant to echo the parent's wording: "And the effect exists and is known for the other planets now; the mass increase is proportional to velocity-squared and the velocity difference depends on the eccentricity of the orbit, so it's an order of magnitude smaller for Earth, but still measurable and now known" With regard to that, here is a table of exactly and approximately solved perihelion advances of all eight planets and Pluto: <https://www.ncbi.nlm.nih.gov/pmc/articles/PMC9635960/table/tab2/ https://www.ncbi.nlm.nih.gov/pmc/articles/PMC9635960/table/t...>. [Sourced from an interesting (but as yet uncited, and weirdly published in TSWJ which let through some minor typos like "elliptik") paper in the category of "the expansion of the universe does not affect solar system orbits": <https://www.ncbi.nlm.nih.gov/pmc/articles/PMC9635960/ https://www.ncbi.nlm.nih.gov/pmc/articles/PMC9635960/> [*]]. Worth noting that Mars's orbit is highly eccentric at 0.0934, and its eccentricity is growing (as is Mercury's). Venus's orbit is much less eccentric at 0.0068. (Greater precision and eccentricities of the other planets can be found at <https://calgary.rasc.ca/orbits.htm https://calgary.rasc.ca/orbits.htm>). [*] Taking the geometric mean across significantly different calculated values of the cosmological constant using their method (which I have no problem with) is to me a mathematical trick too far. That averaging does arrive at the conventional order of magnitude of \Lambda in m^-s, but I wonder if that's coincidence. The first author is essentially a mathematician <https://orcid.org/0000-0001-9343-6062 https://orcid.org/0000-0001-9343-6062>, and whether the metric for the solar system must have a \Lambda term is much more a physical argument. On that front, perhaps the authors should have cited another Iorio paper along with their [10], namely "Can solar system observations tell us something about the cosmological constant?" (2006) < https://scholar.google.com/citations?view_op=view_citation&hl=en&user=rt8ooIEAAAAJ&citation_for_view=rt8ooIEAAAAJ:W7OEmFMy1HYC https://scholar.google.com/citations?view_op=view_citation&h...> although tbf Iorio cites his earlier and more popular paper in [10].