4 ms·

And since readers may find this interesting: The various writeups all say that relativity explained the precession of Mercury's orbit, but they never say exactl

And since readers may find this interesting: The various writeups all say that relativity explained the precession of Mercury's orbit, but they never say exactly what factor was accounted for by relativity. The answer: Mercury is heavier when it is at perihelion in its elliptical orbit, because it's moving faster. The increased relativistic mass makes for increased momentum at perihelion, which carries the planet a little farther than expected before it starts to swing back up out of the gravity well, so the perihelion precesses more. 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. It's also known for other objects: one example is stars in elliptical orbits around the galaxy's central black hole, which undergo extra precession in the same way.

sigmoid10 3y ago
This might sound true if you have a superficial understanding of relativity (and I would not be surprised to see people repeating this stuff on the internet), but it's completely false. The correction to Mercury's orbit precession from General Relativity has nothing to do with Mercury itself (its mass, speed or otherwise). It is purely based on the sun's gravitational field. If you want to interpret it as a potential field in the classical Newtonian sense, it basically picks up an additional 1/r^3 term. That's what disturbs the orbit and it's also the reason why this effect is negligible for other planets apart from Mercury, because it drops off much faster than the normal 1/r potential. Any object very close to a large mass like the sun would feel this distortion, irrespective of how fast it travels.
- antognini 3y ago
  Yes, this is correct. The orbital precession occurs even for a point mass.
  - plank 3y ago
    Former physicist myself: Not sure I understand your (and parents) argument. Given the Suns gravitational field, and the orbit that Mercurius has, the speed of the planet can be determined. So the argument that a point mass would have the same precession does not dispute the argument that it is the relativistic mass of the point mass/planet that determines the precession. I will not venture into arguments* whether grandparents explanation is ‘correct’, ‘best’ or ‘useful’, there is an equivalence between ‘gravitational field + orbit’ being the reason and ‘(relativistic) speed + orbit’ being the reason. * A sybling comment states that relativistic mass is avoided in modern physics. As someone who did physics 30 years ago, I can not deny or corroborate this statement. And indeed, the explanation of grandparent is not the way I myself think about the perihelium precession. But that does not make me certain enough to say it is ‘wrong’. Edit: typo ‘of’ —> ‘or’
    - sigmoid10 3y ago
      Just think about it. If relativistic mass (a term that indeed tends to be avoided by actual physicists for good reasons, because it confuses concepts of mass and energy which are clearly defined in relativity) was the explanation, then the shape of the planet's orbit itself would influence the anomalous precession component from General Relativity. And you would see no relativistic effect at all for perfectly circular orbits with zero eccentricity. That is simply not the case.
      - plank 3y ago
        The explanation of grandparents for the perihelium effect of Mercury tried to explain the 'overshoot'. And yes, in a perfectly circular orbit there would be no overshoot?? I do not think grandparent tried to say that all relativistic effects were caused by the increase of the 'relativistic' mass, just that the overshoot was. So I am sorry: I cannot see how this argument disproves the explanation of grandparent. And just to be clear: I do not think in the way of grandparents argument myself. Kind regards, Roel
        sigmoid10 3y ago
        >And yes, in a perfectly circular orbit there would be no overshoot?? That's where you and the other comment are wrong. The only way to get a stable circular two body orbit is for newtonian potentials which are exactly 1/r shaped. Add any other term (like 1/r^3 for General Relativity) and you essentially get more or less chaotic movement over time, similar to the three body problem in newtonian gravity. See Betrand's theorem [1]. [1] https://en.wikipedia.org/wiki/Bertrand%27s_theorem https://en.wikipedia.org/wiki/Bertrand%27s_theorem
kmm 3y ago
That doesn't sound right. Relativistic mass as a concept is avoided in modern physics because it doesn't yield much insight and is hard to keep consistent. Involving gravity only makes it worse, because it's very hard to consistently include gravity in special relativity. For one, does the relativistic mass gravitate or not? And if you actually do go through the calculations, you find that you do not get the observed result. This paper[0] sums up a few of them. 0 (PDF): http://kirkmcd.princeton.edu/examples/perihelion.pdf http://kirkmcd.princeton.edu/examples/perihelion.pdf
- raattgift 3y ago
  Wow, even for a Kirk McDonald "pedagogic diversion", there is a lot in footnote 4. There are scattered typos, but it's an excellent annotated bibliography. I was unaware of some of the 21st century papers it lists. A small thing, and it could just be me. I'm not sure where McDonald is going with respect to his point in footnote 7 about his reference [54] (Einstein 1911, https://onlinelibrary.wiley.com/doi/abs/10.1002/andp.19113401005 https://onlinelibrary.wiley.com/doi/abs/10.1002/andp.1911340... (German), https://einsteinpapers.press.princeton.edu/vol3-trans/393 https://einsteinpapers.press.princeton.edu/vol3-trans/393 (English)) and unfortunately his ...kirkmcd.princeton.edu/examples/GR/... links all want passwords. It's obvious that the footnote is about §2 of [54], but to me it does not really engage with the \gamma m in the text referring to footnote 7 (or indeed the content of your own comment's parent comment). Instead the (spring-balance-indicated, i.e. "contact"[1]) change of weight when light is beamed from one weighed system to another is diagnostic, with the focus of thought being on how that beamed light is adapted when acceleration or gravitation is relevant. This is further developed into gravitational redshift in [54]§3 (which, frankly, is an amazing bit of Einstein's early thinking). So I don't see how footnote 7 fits the perihelion precession topic, where the (rest) masses are defined as constant and radiation pressure is ignored. [The sodium quasi-cometary tail of Mercury is cool, though. <https://www.nature.com/articles/s41467-020-18220-2 https://www.nature.com/articles/s41467-020-18220-2>] [1] https://math.ucr.edu/home/baez/physics/General/Weight/whatIsWeight.html https://math.ucr.edu/home/baez/physics/General/Weight/whatIs...
canjobear 3y ago
If that’s true then it should be possible to predict the precession from special relativity alone.
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?

sigmoid10 3y ago

This might sound true if you have a superficial understanding of relativity (and I would not be surprised to see people repeating this stuff on the internet), but it's completely false. The correction to Mercury's orbit precession from General Relativity has nothing to do with Mercury itself (its mass, speed or otherwise). It is purely based on the sun's gravitational field. If you want to interpret it as a potential field in the classical Newtonian sense, it basically picks up an additional 1/r^3 term. That's what disturbs the orbit and it's also the reason why this effect is negligible for other planets apart from Mercury, because it drops off much faster than the normal 1/r potential. Any object very close to a large mass like the sun would feel this distortion, irrespective of how fast it travels.

antognini 3y ago
Yes, this is correct. The orbital precession occurs even for a point mass.
- plank 3y ago
  Former physicist myself: Not sure I understand your (and parents) argument. Given the Suns gravitational field, and the orbit that Mercurius has, the speed of the planet can be determined. So the argument that a point mass would have the same precession does not dispute the argument that it is the relativistic mass of the point mass/planet that determines the precession. I will not venture into arguments* whether grandparents explanation is ‘correct’, ‘best’ or ‘useful’, there is an equivalence between ‘gravitational field + orbit’ being the reason and ‘(relativistic) speed + orbit’ being the reason. * A sybling comment states that relativistic mass is avoided in modern physics. As someone who did physics 30 years ago, I can not deny or corroborate this statement. And indeed, the explanation of grandparent is not the way I myself think about the perihelium precession. But that does not make me certain enough to say it is ‘wrong’. Edit: typo ‘of’ —> ‘or’
  - sigmoid10 3y ago
    Just think about it. If relativistic mass (a term that indeed tends to be avoided by actual physicists for good reasons, because it confuses concepts of mass and energy which are clearly defined in relativity) was the explanation, then the shape of the planet's orbit itself would influence the anomalous precession component from General Relativity. And you would see no relativistic effect at all for perfectly circular orbits with zero eccentricity. That is simply not the case.
    - plank 3y ago
      The explanation of grandparents for the perihelium effect of Mercury tried to explain the 'overshoot'. And yes, in a perfectly circular orbit there would be no overshoot?? I do not think grandparent tried to say that all relativistic effects were caused by the increase of the 'relativistic' mass, just that the overshoot was. So I am sorry: I cannot see how this argument disproves the explanation of grandparent. And just to be clear: I do not think in the way of grandparents argument myself. Kind regards, Roel
      - sigmoid10 3y ago
        >And yes, in a perfectly circular orbit there would be no overshoot?? That's where you and the other comment are wrong. The only way to get a stable circular two body orbit is for newtonian potentials which are exactly 1/r shaped. Add any other term (like 1/r^3 for General Relativity) and you essentially get more or less chaotic movement over time, similar to the three body problem in newtonian gravity. See Betrand's theorem [1]. [1] https://en.wikipedia.org/wiki/Bertrand%27s_theorem https://en.wikipedia.org/wiki/Bertrand%27s_theorem

kmm 3y ago

That doesn't sound right. Relativistic mass as a concept is avoided in modern physics because it doesn't yield much insight and is hard to keep consistent. Involving gravity only makes it worse, because it's very hard to consistently include gravity in special relativity. For one, does the relativistic mass gravitate or not? And if you actually do go through the calculations, you find that you do not get the observed result. This paper[0] sums up a few of them. 0 (PDF): http://kirkmcd.princeton.edu/examples/perihelion.pdf http://kirkmcd.princeton.edu/examples/perihelion.pdf

raattgift 3y ago
Wow, even for a Kirk McDonald "pedagogic diversion", there is a lot in footnote 4. There are scattered typos, but it's an excellent annotated bibliography. I was unaware of some of the 21st century papers it lists. A small thing, and it could just be me. I'm not sure where McDonald is going with respect to his point in footnote 7 about his reference [54] (Einstein 1911, https://onlinelibrary.wiley.com/doi/abs/10.1002/andp.19113401005 https://onlinelibrary.wiley.com/doi/abs/10.1002/andp.1911340... (German), https://einsteinpapers.press.princeton.edu/vol3-trans/393 https://einsteinpapers.press.princeton.edu/vol3-trans/393 (English)) and unfortunately his ...kirkmcd.princeton.edu/examples/GR/... links all want passwords. It's obvious that the footnote is about §2 of [54], but to me it does not really engage with the \gamma m in the text referring to footnote 7 (or indeed the content of your own comment's parent comment). Instead the (spring-balance-indicated, i.e. "contact"[1]) change of weight when light is beamed from one weighed system to another is diagnostic, with the focus of thought being on how that beamed light is adapted when acceleration or gravitation is relevant. This is further developed into gravitational redshift in [54]§3 (which, frankly, is an amazing bit of Einstein's early thinking). So I don't see how footnote 7 fits the perihelion precession topic, where the (rest) masses are defined as constant and radiation pressure is ignored. [The sodium quasi-cometary tail of Mercury is cool, though. <https://www.nature.com/articles/s41467-020-18220-2 https://www.nature.com/articles/s41467-020-18220-2>] [1] https://math.ucr.edu/home/baez/physics/General/Weight/whatIsWeight.html https://math.ucr.edu/home/baez/physics/General/Weight/whatIs...

canjobear 3y ago

If that’s true then it should be possible to predict the precession from special relativity alone.

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?