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Geodetic effect

  1. Feb 21, 2012 #1
    I recently read in astronomy magazine about Gravity prob b and it findings, Can someone please explain Geodetic effect and how it proves relativity?

    Thanks
    Bearded
     
  2. jcsd
  3. Feb 21, 2012 #2

    Matterwave

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    Basically, GR says that because space-time is curved (due to the presence of the Earth), if we move a vector (i.e. the angular momentum vector of a spinning gyroscope) around in a closed loop, the vector will not return to it's original orientation, and will tend to have "tipped over" in some sense. This is the geodetic effect basically, and gravity probe b found that the spinning gyroscope did indeed change it's axis of rotation by a very little bit after going around the Earth many times.
     
  4. Feb 22, 2012 #3

    A.T.

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  5. Feb 22, 2012 #4

    Bill_K

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    A.T., I'm sure your diagram is well-intentioned, but it is absolutely wrong, and anyone who has seen it is hereby requested to purge it from their minds!

    It is not true in any sense that the spacetime curvature surrounding a massive object is analogous to a cone. While drawing a cone makes it easy to visualize, it is completely wrong. A cone is formed from a flat piece of paper by removing a section and pasting the remaining part together. What you are left with has an angular defect: its circumference is less than two pi. And sure enough, a vector transported around the circumference will fail to come back to its original position for that reason. Such spacetimes are quite anomalous, although they have been considered for example in cosmic strings.

    Schwarzschild does not have this property. The surfaces r = const in Schwarzschild are perfectly normal two-spheres. A vector carried around a great circle on this two-sphere will come back to its original position.

    See the Wikipedia article for a valid derivation of the geodesic, aka deSitter, effect.
     
  6. Feb 23, 2012 #5

    A.T.

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    In that case, you should also urge Stanford and Kip Thorne to stop using it to explain the geodetic effect:
    http://einstein.stanford.edu/SPACETIME/spacetime4.html#geodetic_effect

    It approximates the spatial geometry Schwarzschild (Flamm's paraboloid) as I explained here:
    https://www.physicsforums.com/showthread.php?p=3586633&postcount=9

    The circumference of a circle around a big mass is also less than 2 pi * proper radius, according to Schwarzschild geometry.

    Flamm's paraboloid doesn't have any angular defect?

    Only in the sense in which the rim of the cone is a "perfectly normal" circle. And yet there is an angular defect on that cone.
     
    Last edited: Feb 23, 2012
  7. Feb 23, 2012 #6

    Bill_K

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    A.T., Thanks for your reply. I remember now the previous thread. But I see you are still mistaken.

    Kip Thorne - It's just as wrong when Kip says it as when you say it. I'll mention it to him.

    Flamm's paraboloid - as you say it models the spatial geometry. Which is unrelated to this effect, since precession depends on a timelike curve, the world line of the observer. The geodesic or deSitter precession is not a new effect in general relativity, merely a generalization of the Thomas precession - a very similar effect in flat space, where there is certainly no conical defect to blame it on. All that's new in general relativity is that it happens for geodesic world lines as well.

    Also, the effect is continuous and local. Your conical explanation makes it out to be global - something that is noticed only after a complete orbit when the initial and final directions fail to coincide. In fact you do not have to go all the way around to see it. And it would happen even if the orbit were not closed. Or if the path were closed and did not encircle the origin. The gyroscope precesses continuously. Its axis is observed to slowly track a path across the sky relative to the distant stars.

    The importance of the effect is to show that the definition of a local nonrotating frame does not coincide with the distant one, and even depends on the motion of the observer.
     
  8. Feb 23, 2012 #7

    bcrowell

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    One expects both a geometrical precession (due to spatial curvature) and a Thomas precession. But when you write down the equations of motion in GR, these do not appear as two separate terms with clearly distinguishable physical interpretations. I have some discussion of this in this book http://www.lightandmatter.com/genrel/ at the end of section 6.2.5.
     
  9. Feb 24, 2012 #8

    A.T.

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    That is how I see it too. The spatial part is easier to visualize (with the cone diagram) hence it's OK to use it, and eventually mention that this explains 2/3 of the effect.

    However, I would prefer a diagram where the vector is within the cone surface. The gyro-orientation is a spatial vector, so it should be within the 2D-surface which represents 3D-space.
     
    Last edited: Feb 24, 2012
  10. Feb 24, 2012 #9

    Bill_K

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    For a particle in a circular orbit about a Schwarzschild mass, with coordinate angular velocity ω = dφ/dt, the angular velocity of the gyroscope axis is Ω = γ2ω(r - 3m)/r. The first part γ2ω can be gotten by setting m = 0, and is the same as the Thomas precession rate for Minkowski space. The second part -3mγ2ω/r may be regarded as an additional deSitter rate. But as you've already mentioned, the derivation yields both terms together so although considering them separately is convenient, it is somewhat artificial. Together they represent the precession rate of a local nonrotating frame, nonrotating in the sense that it is Fermi-Walker transported along the particle's worldline, whatever that may be.

    I quote the result to point out that one cannot claim a constant ratio between the two terms like "2/3", since they have different radial dependence. Naturally the deSitter rate goes to zero at r = ∞ while the Thomas term does not. Also, they have opposite signs and hence represent precessions in opposite directions. The Thomas term is retrograde while the deSitter term is prograde. At the radius of a photon orbit r = 3m they cancel.
     
  11. Feb 24, 2012 #10
    Is there a writeup (for people who know GR) of this claim that 2/3 of the effect is due the non-Euclidean relationship between radius and circumference?

    I have to agree with Bill K that fundamentally the effect is local, and the diagram is pretty misleading in that sense. But I'm sure there's something behind it, if Kip is using it. It's tough to know how much to simplify when giving popular explanations.
     
  12. Feb 24, 2012 #11

    A.T.

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    I was referring to the link posted by bcrowell, which deals with Gravity Probe B:
    http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 [Broken]
    scroll to 6.2.5
     
    Last edited by a moderator: May 5, 2017
  13. Feb 24, 2012 #12

    bcrowell

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    Two contradictory interpretations:

    Misner, Thorne, and Wheeler, Gravitation, p. 1118
    Rindler, Essential Relativity, 1969, p. 141

    I don't have the books handy, but IIRC Rindler does make the 2/3 claim. The whole thing is just a heuristic/popularization.
     
  14. Feb 28, 2012 #13

    Bill_K

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    It's amazing how many different answers have been given to this problem! Some of them approximate, others unsupported, others simply incorrect. The "2/3" it seems, comes from comparing a (wrong) derivation using only the spatial geometry to a second derivation done right. Wikipedia's article pulls their result directly from a set of rotating coordinates, getting another wrong answer. Kip's discussion in MTW p1118 is presumably correct but is rather hard to follow and assumes the PPN approximation. I can't find anywhere a clean treatment. And the results are really quite simple and nice, so there they are.

    The Kerr metric confined to the equatorial plane is ds2 = (r-2m)/r dt2 + 4ma/r dφdt - (r2+a2+2ma2/r) dφ2 - r22 dr2 where Δ ≡ r2 - 2mr + a2.

    For a particle moving with coordinate angular velocity ω = dφ/dt we have γ2 = 1/(1 - (r2+a22 -2m/r(1-aω)2)) and radial acceleration A = Δγ2[m(1-aω)2 - r3ω2]/r4. For a particle in orbit we would have geodesic motion, A = 0. You'll notice that this condition is a quadratic, generalizing Kepler's law and implying two different roots for ω, one for revolution in the same sense as the Kerr rotation, the other for the opposite sense.

    Now if you have a gyroscope attached to the particle, with spin given by a spacelike unit vector S, the components of S will vary sinusoidally, S ~ sin Ωt. If Ω = ω, the spin axis would point in a fixed direction in space and there would be no precession. So the difference between Ω and ω measures the precession. Here's the really beautiful result:

    Ω = γ2ω - 3γ2ωm(1-aω)/r + γ2ma(1-aω)2/r3

    The first term is the Thomas precession, the second is deSitter, and the third term is the frame dragging. Again this formula covers all the bases - you can restrict it to orbital motion if you want, or you can just set ω = 0 and get the effect for a particle sitting at rest on the Earth's surface.
     
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