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Geodetic precession and parallel transport

  1. Jul 1, 2010 #1
    Could anyone give me a descriptive picture on WHY geodetic precession occurs? I understand the equations from which it follows, so I can derive it algebraically, but I would like to get an intuitive feeling of why it occurs too.

    My problem is the following: parallel transport of vectors along a curved line (i.e., NON-geodesic!) results in the vector changing its orientation with respect to the line. Examples: (1) vector on a 2D Euclidean plane parallel transported along a curve; (2) vector parallel transported along a line of latitude (other than the Equator!) on the spherical surface of Earth. [This is the reason why Foucault's pendulum changes its orientation with respect to the laboratory on Earth (except on the Equator) as it rotates through 2Pi in a day.]

    However, as is also well known, if the vector is parallel transported along a GEODESIC line (e.g. straight line on a 2D Euclidean plane or the Equator on the spherical surface of Earth), it KEEPS its orientation with respect to the line along which it was parallel transported.

    I am confused, then, as to WHY the orientation of a gyroscope which is orbiting Earth (i.e. floating FREELY, i.e. moving along a GEODESIC) changes. Could anyone help me out? As I said, I'm looking for an intuitive answer (preferably using a geometric picture), rather than the algebraic derivation.
     
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  3. Jul 1, 2010 #2

    Cleonis

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    Check out http://arxiv.org/abs/gr-qc/0411060" [Broken] There you may well find clues that help you.

    Max Tegmark uses the river model in his lectures.
    http://mitupv.mit.edu/wp/attach/4581/barry.pdf"


    As Andrew Hamilton emphasizes, the river model can't be taken literally, but it's a good heuristic. Anything that is visualized in terms of the river model will match the predictions in terms of spacetime curvature.

    If I remember correctly there is mention of geodetic precession in the discussion by Andrew Hamilton.



    You write about attempts to frame an analogy with parallel transport over the surface of a sphere. I think that analogy doesn't apply.

    Let me set up some background:
    Generally a setup to show geodetic precession has a vectorial object move along some loop trajectory. When the loop is closed the geodetic precession becomes clear.
    The case of parallel transport along the surface of a sphere is a special case in the sense that there is a subclass of trajectories (motion along great circles of the sphere) with the property that no geodetic precession occurs.
    It is my understanding that the curved spacetime as described by GR does not have such a subclass of geodeticprecession-free trajectories. The geodetic precession will always occur, regardless of whether the trajectory is along a geodetic or not.

    I hope that in the discussions by Andrew Hamilton and Max Tegmark you will find the clues that you need to answer your question.

    [later edit]
    I've done some rereading, and I need to correct myself.
    The gravity Probe B experiment data enable research on two kinds of relativistic effects: geodetic precession and frame dragging. The information in the Andrew Hamilton 'River model' article is relevant only for the frame dragging effect, I think.

    The difficulty, it seems, is not so much understanding of geodetic precession itself. It seems that the cause of the confusion is an unfounded assumption that the case of parallel transport over the surface of a sphere is a suitable analogy.

    How about the analogy of parallel transport along the inner surface of a cone? (The kind of cone that is a visual analog to a gravity potential well) I think that in that analogy every trajectory (along a geodetic or not) will feature geodetic precession.
    [end later edit]



    By the way, I am aware that some authors mention the Foucault pendulum as an example of an effect arising from parallel transport along a curved surface. That analogy kind of works as a approximation, but only with limited scope. For example, the approximation holds only if there are many swings of the pendulum for each cycle around the trajectory. The 'parallel transport' view cannot be pushed to high accuracy, because it doesn't quite fit. The description of the Foucault pendulum precession in terms of classical mechanics is more general; it can be pushed to any desired level of accuracy.
    I discuss those things in the http://www.cleonis.nl/physics/phys256/foucault_pendulum.php" [Broken] on my website.

    Cleonis
     
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  4. Jul 1, 2010 #3
    Cleonis:

    Thanks a lot for your suggestion. Although I couldn't find direct references to geodetic precession in the River Model links you mentioned, but I am very hopeful that I'll find the answer to my problem there, so I'll check those papers in detail.

    I am a little doubtful of your discussion of GR curved spacetime vs. a spherical surface in 3D Euclidean space. I think that - at least on 2D surfaces in 3D Euclidean space - no precession should occur along ANY geodesic on ANY curved surface (not just along great circles on a sphere), so it seems very strange to me why 4D curved spacetime should behave otherwise. After all, the DEFINITION of geodesic is the same in GR curved spacetime as on 2D curved surfaces: lines (or worldlines) along which, if the tangent vector to the line is parallel transported, it will remain a tangent vector.

    In fact, as is discussed in some GR texts, the orientation of the gyroscope does NOT precess with respect to the local INERTIAL frame (it points along the same direction in the spaceship in which the gyroscope travels), so in this sense indeed NO precession occurs along the geodesic. But somehow, as viewed relative to a well-defined external reference direction - e.g the Earth-Sun radial line - precession does occur. This is, basically, what is hazy to me. In any case, I'll check out what the river model says in this regard.

    BTW, your Foucault pendulum website, of which I had been aware before, is truly great!

    Thanks again,
    nnn_bbb
     
  5. Jul 1, 2010 #4

    Mentz114

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  6. Jul 1, 2010 #5

    Cleonis

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    As I state in my later edit to post #2 in this thread, I think there is an example of a curved 2D surface embedded in a 3D space where all trajectories, regardless of whether they are geodesics or non-geodesics, will feature geodesic precession. That example is the inverted cone that is so often presented as a visualization of a gravity potential well.

    I have no proof though; I'm reasoning by analogy here.
    - When a vector is parallel transported along a great circle of a sphere then that is equivalent to parallel transport along the shortest planar ring that surrounds a cylinder. In that case there is no geodetic precession because there is no curvature.
    - But in the case of the inverted cone there is no loop-closing trajectory that is equivalent to going around a cylinder. All loop-closing trajectories are affected by the curvature
     
  7. Jul 1, 2010 #6

    bcrowell

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    I don't think this is right. I think parallel transport along a geodesic preserves the tangent vector, but not other vectors.

    For (my own) calculation of the effect, see http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 [Broken] , subsection 6.2.5. The differential equations for parallel transport give an eigenvalue problem, and one of the eigenfrequencies is zero. As discussed in the link, the reason we get one eigenfrequency of zero is that a vector tangent to the geodesic doesn't precess. The other two nonzero eigenfrequencies affect vectors that are not tangent to the geodesic.
     
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  8. Jul 2, 2010 #7
    But on a cone, unlike on a sphere, you cannot construct a loop-closing geodesic! So it seems obvious to me why all loop-closing trajectories have precession: because they are NOT geodesic! Along (open) geodesics on a cone precession does NOT occur.

    Incidentally, I think this whole problem is not really connected to loop-closing vs. not loop-closing geodesics. I still strongly feel (by analogy) that precession shouldn't occur along ANY geodesic on ANY manifold. [By the way, the geodesic involved in the GP-B experiment is not a loop-closing geodesic anyway, since we don't go back to the same TIME instant after one cycle.]
     
  9. Jul 2, 2010 #8
    This sounds very interesting. But could you give me a geometric analogy with a precessing (non-tangent) vector on a 2D curved surface in 3D euclidean space? Or is the quest for such an analogy hopeless??
     
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  10. Jul 2, 2010 #9
    This link is very interesting too, and might take me closer to the understanding. Thanks!
     
  11. Jul 2, 2010 #10

    bcrowell

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    I don't think it works in 2 dimensions. In 2 dimensions, you can carry two gyroscopes at right angles to one another. If one of them is tangent to the geodesic you're following, and it stays tangent, then there is no way for the other one to change its direction relative to the geodesic while maintaining its 90-degree angle with the first vector.
     
  12. Jul 2, 2010 #11

    Cleonis

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    To address that, let me link to an image available at wikimedia commons:
    http://commons.wikimedia.org/wiki/File:GravityPotential.jpg" [Broken]

    I refer to the 2D surface depicted in that image as an 'inverted cone'. It represents a gravitational potential well.

    In this analogy orbital motion is like the motion of an object that slides frictionless over that surface, subject to a force down the gradient of the slope.

    (This picture can also be adapted to depict the river model of GR spacetime curvature. Imagine a flow over that surface, with the property that through all concentric circles there is the same total flux. Hence the closer to the lowest potential the faster the flow. To make the model work it is stipulated that an object on that surface is affected exclusively by the acceleration of the flow. This accounts for the centripetal acceleration of orbiting objects.)

    An orbit is a loop-closing geodesic.
     
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  13. Jul 2, 2010 #12
    Perhaps I can offer an insight into one small piece of the puzzle. One component of the geodetic precession is called Thomas precession. This is a kinematic effect that can be explained by Special Relativity. Consider an extended rocket in flat space that has inertial motion parallel to its length, relative to a given inertial reference frame S. The rocket has a booster rocket located exactly at its centre of mass that is at right angles to the length of the rocket. Every so often the booster is fired. In Newtonian terms we would expect the rocket to be displaced sideways but still be orientated exactly parallel to its original direction. In SR the impulse spreads out from the centre to the nose and tail of the rocket simultaneously in the frame of the observers inside the rocket, but to the external observers at rest in frame S, the sideways impulse does not happen simultaneously and the tail is boosted sideways slightly earlier than the nose. The end result is that the rocket not only moves sideways but is given a slight rotation away from its original heading from the point of view of the S frame observers. In the gravitational context the acceleration acts on the centre of the mass of the extended orbiting object and presumably a similar effect occurs due to the lack of simultaneity in the non rotating frame. I have not studied this in great depth and this is just my superficial understanding of Thomas precession, but I hope it is helpful. I welcome any corrections or further insights from other contributers.
     
  14. Jul 3, 2010 #13

    bcrowell

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    IMO it doesn't work well if you try to interpret part of the geodetic effect as a Thomas precession. For a discussion of this, see http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 [Broken] , at "One will see apparently contradictory statements in the literature..." For one thing, the sign of the effect is wrong.
     
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  15. Jul 3, 2010 #14
    Hi Ben,

    I respect your opinion on this because as I said earlier this is a topic I have not yet gone into it in any detail. I got the impression that Thomas precession is part of the geodetic precession from this Wikipedia article http://en.wikipedia.org/wiki/Thomas_precession
    The link associated with the phrase "de Sitter precession" in the article leads to http://en.wikipedia.org/wiki/De_Sitter_precession which says
    I put the two statements together and came to the conclusion that in the curved spacetime of general relativity, Thomas precession combines with a geometric effect to produce the geodedetic effect. However, this is new material to me and I hope to learn more as this thread progresses.
     
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  16. Jul 3, 2010 #15

    bcrowell

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    If you look at the link I gave (to my own book), you'll see quotes from well known relativists that superificially contradict one another on this issue. After that there's a more detailed discussion of why the question is a complicated one.
     
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