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Sun forms different geodesic

  1. May 15, 2007 #1
    Earth follws a straight path in 4-d space time.ok.now the earth moves over the geodesic formed by the sun's gravity.now we also have other 7 planets.So does it mean that the sun forms different geodesic for differnt planets. if my question does make some logic than please explain me.
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  3. May 15, 2007 #2


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    The geodesics are determined by the mass, energy and momentum distribution of the sun and the planets. Since the sun is the most massive by far, it has more effect than the planetary masses. I'm not sure what you're asking, though.
  4. May 15, 2007 #3


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    This is discussed for instance in MTW's textbook, 'Gravitation', on pg 1126. "Do the planets and sun move on geodesics?".

    The short answer is "not quite, but the approximation is very good for GR".

    The longer answer is that answering the question first requires setting up an approximation scheme.

    This is a subtle but interesting point. The earth, not being a test particle, contributes to the space-time geometry. The background geometry, the geometry that would exist without the Earth being present, doesn't actually exist.

    It may not be obvious if one is not familiar with GR, but the coupling being talked about is just the "spin-orbit" coupling between the Earth and the moon, i.e. it is the torque caused by the moon on the tidal bulges ('multipole moments') of the Earth. Because the Earth is experiencing a torque, it's not a free body. i.e. not following a geodesic.

    Not mentioned in this section of MTW, because it does not follow from linearized theory, is gravitational radiation, another mechanism that could be said to cause actual bodies to depart from geodesics.

    Note that in some OTHER theories of gravity, most notably Brans-Dicke theories, geodesic motion may not be as good an approximation as it is in GR. Google for instance the "Nordtvedt effect". Which is actually a non-effect, because while it is predicted by some alternate theories of gravity it has not actually been observed.
  5. May 15, 2007 #4
    That was at a higher level than the question I thought the OP was asking (for which's purposes, the planets basically do follow geodesics of the space-time curved by the sun). Guess that means further clarification would be good, Milind.

    Basically, no, the path of a planet doesn't depend on, say, it's mass etc (ie. "which planet it is"). But (on the other hand) yes, like drawing straight lines on a piece of paper, an infinite number of different lines (ie. geodesics) could be drawn, depending on where the line starts and the slope/direction of each line (ie. the velocity of the planet, at a particular point in space-time). Another popular analogy is ants walking in straight paths on an apple.. inumerable different paths are possible depending on how the ant begins walking, but the paths do not depend on the size, weight, colour or name of the individual ants.

    You might think of Earth travelling in one straight line, and Jupiter travelling on a different (sortof parallel) straight line. If Jupiter were magically "swapped" with Earth, Earth would just continue along Jupiter's geodesic (despite the difference in mass) and likewise Jupiter along Earth's geodesic (complete with 365 day year etc). However, If instead we just made Jupiter magically go faster suddenly, then that would alter its path, and it would continue along a new different geodesic. Likewise, if Earth were moved to a completely different vacant position in space-time, it would follow a completely new path.
    Last edited: May 15, 2007
  6. May 15, 2007 #5
    I just read the relevant MTW pages and I must say that I find them highly dissatisfying, especially for a textbook.

    Since an extended object like the earth or the moon does not follow a world line or not even a world tube the question whether it follows a geodesic is really a question about every individual particle of this extended object.

    An extended object is a bundle of world lines. I imagine a bundle of slightly twirling cables held together by clips of a typical Dolby surround system here. :wink:

    If any of the world lines in this bundle is influenced by electro-magnetic, weak or strong forces it is not free falling. So in that respect the earth or any other planet is obviously not traveling on a geodesic.

    But even if the object would be some kind of dust would every fermion travel on a geodesic or only bosons? And what if the constituents are not point particles?

    Is the answer in the trace term?
    Last edited: May 15, 2007
  7. May 15, 2007 #6
    OT, but clearly the "center" of an extended object does follow a world line, so is this just semantics? Also, how is a "world tube" different to a (continuous) "bundle of world lines"?
    Regards MTW, the question was important historically because GTR was largely motivated by the idea that geodesics would model planetary motion better than Keplerian orbits, so obviously it could have been "catastropic" if GTR predicted planets deviating from geodesics non-negligibly.
    Yeah, does intrinsic spin couple to the space-time? I'd expect so, but I've been warned against presuming it. For that matter, are elementary particles even pointlike in that limit? Guess quantum gravity experiments aren't easy to perform though..
    Last edited: May 15, 2007
  8. May 15, 2007 #7


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    Scholarly replies, indeed, but I reckon the OP doesn't know what a geodesic is and is overwhelmed. I was waiting for this happen. I (affectionately) call you guys the Relativity Police. Long may your fibres bundle.
  9. May 15, 2007 #8
    Nice, but how do you calculate a center in a collection of converging and diverging worldlines? I would venture there is no center not even a theoretical one. :smile:

    Indeed OT, but briefly, a bundle of world lines is topologically very different from a tube.
  10. May 15, 2007 #9


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    Do you just mean that the Earth does not follow the path that would be a geodesic for a test particle in a spacetime curved by the sun alone, or that the Earth does not even follow a geodesic in the actual spacetime whose curvature includes a contribution from the Earth? In other words, if a massless test particle were placed at the center of the Earth with the same instantaneous velocity, then assuming the particle was neutral to all other forces (so it could move through the body of the Earth like a neutrino), would the geodesic-following test particle remain lined up with the Earth's center or would it diverge from it over time?
  11. May 15, 2007 #10
    To be precise the Earth follows a geodesic which is not a straight line in all cases but the straightest possible line, such as a great circle on a sphere.

    All the planets form a system which is described by Einstein's equations. A geodesic is what a test particle follows where as test particle has no spatial dimensions (and thus no spin) and has a mass which is neglegible for being a test particle (i.e. it doesn't alter the spacetime significantly).

    Different planets are already on different geodesics because they travel at differing speeds at differing points and directions in space. If you want an exact situation then the stress-energy-momentum tensor will be time dependant as will the spacetime and would be a nightmare to compute by hand. You'd need some strong experience in numerical methods for differential geometry to solve it on a computer.

  12. May 15, 2007 #11
    Empirical verification!

    Interestingly, no exact solution can ever be empirically verified, not even in principle! Any measurement of the solution will disturb it non-linearly. One would need to know the initial conditions of the disturbance to include it in the solution, and that simply brings the measurement problem one step back, ad infinitum. :smile:
  13. May 15, 2007 #12
    lol. (You might note my response to the OP, but "we fuzz" should perhaps note the similarity between "policing", "patrolling" and just "trolling".)
    If the Earth's core wasn't liquid, surely the "atom at the middle" would suffice.
    Regards the former: 1) even in Newtonian gravity, an extended body follows a slightly different path compared to if it were a point particle. 2) Even a point particle technically deviates slightly from geodesic motion if it has angular momentum.
    So philosophical; you should have found an exact solution that incorporates the measurement apparatus explicitly. :biggrin:
  14. May 15, 2007 #13


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    A non-spinning test particle at the center of the Earth and the center of the Earth itself would probably diverge because of the interaction of the spin of the Earth with "gravitomagnetic" components of the solar system's gravitational field, wihile the test particle, not having a spin, would follow a different path.

    I believe the path of spinning test particle would be modelled by the papapetrou equations.

    Other than knowing that the effect is very very small, I haven't done much with the Papapetrou equations, though.
  15. May 15, 2007 #14
    I mean to say that.Earth moves on the geodesic formed by the sun's gravity.So along with earth we also have different planets revolving the sun.So does it means that the sun forms different geodesics for different planets.because all the planets are not in a single orbit they are in different orbits
  16. May 15, 2007 #15

    Chris Hillman

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    Does this help?

    When I saw your initial post, I started to reply to this effect:

    Gtr is a nonlinear theory, so it is very difficult to find exact solutions analogous to the exact solution to the two-body found by Newton for his non-relativistic theory of gravitation, much less the n-body problem, which even in Newtonian theory is very complicated. However, one can run numerical simulations and if one knows what one is doing, these will be accurate over specified time scales. You can visualize the result as a rather complicated spacetime (four dimensional Riemannian manifold) with n timelike geodesic paths representing the world lines of the Sun and planets.

    But as pervect and others mentioned, it is actually a nontrivial matter to show that the EFE implies something which is much easier to take as an extra postulate analogous to the Lorentz force law in electromagnetism: that unaccelerated particles have world lines which are timelike geodesics. It is quite remarkable that the field equation of gtr implies the equation of motion of bodies like planets (if idealized as something like point masses--- this isn't quite right, but never mind), since this is not true in Maxwell's theory!
  17. May 15, 2007 #16

    Hi pervect
    I know that true geodesic motion is ideolised for test particles.But many a times we can consider it for real mass.I accept that due to this some discrepencies arise likethe real mass affects the background geometry.But here i would just like to know as why different planets revolve in different orbits ie why different planets move along different geodesics
  18. May 15, 2007 #17
    pete.If i am not wrong than In general relativity, gravity is not a force but is instead a curved spacetime geometry where the source of curvature is the stress-energy tensor.Now than does it mean that if planets are swapped, as caesium said,for eg if jupiter comes on earths place and earth on jupiter's place than jupiter would keep on moving on earths orbit and earth would keep on moving in jupiter's orbit considering that velocity of jupiter turns to the velocity of earth at earths orbit and earth's velocity changes to jupiter's velocity at jupiter's orbit.
  19. May 16, 2007 #18


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    If you swapped the planets, you would have a different energy-momemtum tensor, so a different solution to Einstein's equation, and a different set of orbits.
    What ? This is meaningless. How could they do otherwise than move in different orbits ? Are we having a language problem here ?

    Are you asking how to calculate the orbits ? You can find that in any book on GR theory.
  20. May 16, 2007 #19
    Do you know what "spacetime curvature" means in real life? It means nothing more and nothing less that tidal forces. The term "gravitational force" is as it is defined to be. I created a web page to illustrate this at
    http://www.geocities.com/physics_world/gr/grav_force.htm - See Eq. (8a).

    In a a non-inertial coordinate system the gravitational force is not everywhere zero.

    The tidal force tensor in NBewtonian Physics is on my web page here

    Compare this with the Riemann tensor here

    Last edited: May 16, 2007
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