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I What is an angle deficit in the curved space in GR?

  1. Jul 4, 2016 #1
    I think this should be equal to the famous precession angle, but with a negative sign:
    ##d\phi = 6\pi m/r##
    correct?
     
  2. jcsd
  3. Jul 4, 2016 #2
    And what is the Gaussian curvature of the space in the GR: ##K=m/r^3## ?
     
  4. Jul 4, 2016 #3
    So, a curvature radius is equal to: ##R=1/\sqrt{|K|}##
     
  5. Jul 4, 2016 #4

    Jonathan Scott

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    It's not exactly clear what you mean by mixing up your "curvature" terms, but the linear curvature of a ruler tangential to the field, relative to typical coordinates, is (at least for weak fields) the same as the Newtonian field converted to units of angle per distance, that is ##g/c^2## where ##g = -Gm/r^2##. This is why something moving tangentially at the speed of light is accelerated twice as fast as something at rest. The total angle for a complete circle with circumference ##2 \pi r## is therefore simply ##2 \pi Gm/rc^2##.

    The usual precession formula depends not only on the curvature with respect to space but also on being in orbit (with the appropriate velocity) and on the second-order term in the time dilation factor, and is a very sensitive test of GR compared with other theories. See for example MTW "Gravitation" section 40.5 and exercise 40.4.
     
  6. Jul 4, 2016 #5
    I think a time has nothing to the geometric curvature of the space alone.
    So, the orbital precession must be equal to the angle deficit along any closed loop in the curved space, and exactly, not other.
     
  7. Jul 4, 2016 #6

    Mentz114

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    If you are referring to planetary orbits in the Schwarzschild vacuum, then the exact solution is in terms of the Weierstrass ##\wp## function.

    See G. V. Kraniotis, S. B. Whitehouse, [/PLAIN] [Broken] http://arxiv.org/abs/astro-ph/0305181[/URL] [Broken]
     
    Last edited by a moderator: May 8, 2017
  8. Jul 4, 2016 #7
    I don't ask about any particular planetary motion model, but about the angular deficit in the GR only!

    If the model can't reproduce the correct angle deficit along the circle, then the rest is nothing but another improvised numerology only!
    df = ?
     
  9. Jul 4, 2016 #8
    m is not as straight forward in General Relativity as it is in Newtonian gravity.

    It seems to me you are looking for generalizations and simplifications that simply do not exist in GR.
     
  10. Jul 4, 2016 #9

    pervect

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    It depends on which plane you select. It's identical to the values of particular components of the Riemann tensor in an orthonormal basis, if that helps. Rather than go into the technical details , I'll give a list of sectional curvatures for the Schwarzschild metric.

    Let ##\hat{t} , \hat{r}, \hat{\theta}, \hat{\phi}## be unit length vectors in the Schwarzschild coordinate system.
    Then in the ##\hat{t}\hat{r}## plane and the ##\hat{\theta}\hat{\phi}## ##\quad K = -\frac{2GM}{r^3} seconds^{-2} = -\frac{2GM}{c^2 r^3} meters^{-2}##
    In the ##\hat{t}\hat{\theta} \quad \hat{t}\hat{\phi} \quad \hat{r}\hat{\theta} \quad \hat{r}\hat{\phi}## planes, ##\quad K = \frac{GM}{r^3} seconds^{-2} = \frac{GM}{c^2 r^3} meters^{-2}##
     
  11. Jul 4, 2016 #10
    A spatial curvature is in the spatial dimensions, thus it's:
    ##K_r\phi=-m/r^3##

    These other curvatures, the mixed like: t-r,, are irrelevant because the GR claims:
    the space is curved, thus we observe the planetary precession... and the light deflection too.

    We observe these all effects in the space directly, not a time is relevant there!
    And even more! The the gravitational redshift is also in the space observed - never in a time!

    So, what is the angular deficit in the GR curved space (gravitationally)?
     
  12. Jul 4, 2016 #11

    Dale

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    This is incorrect. GR claims that spacetime is curved in general. Furthermore, the foliation into space and time is not unique.

    In the Schwarzschild spacetime the traditional foliation and it's spatial curvature is insufficient to account for the observed light deflection. Here is a good page on the topic although it approaches it from the other side (starting with the "time" curvature only and then adding the spatial part)

    http://mathpages.com/rr/s8-09/8-09.htm
     
  13. Jul 4, 2016 #12

    PeterDonis

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    There is no such thing as "the" GR curved space. GR can describe an infinite number of different curved spacetimes, each one corresponding to a different solution of the Einstein Field Equation. So your question is not well-defined. You need to decide which particular curved spacetime you are interested in. Then, as others have pointed out, you need to decide what particular "angle deficit" in that curved spacetime you are interested in.
     
  14. Jul 5, 2016 #13
    Actually, the orbit of a free fall test particle is a geodesic of spacetime, and if you parametrise these geodesics by proper time, then their defining characteristic is that they are world lines which extremise total proper time between given events ( = the principle of extremal ageing ). As such, time is very relevant to this question indeed.
     
  15. Jul 5, 2016 #14

    A.T.

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    Maybe the question is this: Would the orientation of a gyroscope precess at the same rate as it's non-circular orbit does?
     
  16. Jul 5, 2016 #15

    Jonathan Scott

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    OK, that's a third type of precession angle!
    1. The orientation of a gyroscope attached to a free-falling object in orbit precesses by an amount which includes the "de Sitter" effect due to the linear curvature of space and also the relativistic "Thomas precession" which is due to the relativity of changing direction.
    2. The angle defect for a set of static rulers around the path of an orbit does not include the Thomas precession, as that is velocity-related.
    3. The precession of the perihelion of an orbit involves other terms, and although it is affected by the linear curvature of space, it is not the same as a static path, as it relates to the difference in period between the oscillation of the radius of the elliptical orbit and the period of the orbit itself.
     
  17. Jul 5, 2016 #16

    PeterDonis

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    I'm not sure I understand. How is this particular angle defect to be measured? And which components of the Riemann tensor do you think it depends on?
     
  18. Jul 6, 2016 #17

    Jonathan Scott

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    Well, call it a set of static protractors. I mean that if you count up the angle that you have turned through (e.g. relative to an isotropic coordinate system) when constructing a static ring around the path of an orbit, then when the ring closes, the angle will differ from 2 pi by approximately that much. This is determined by the radial gradient of the tangential spatial factor in the metric, the same term that causes the deflection of a tangential light beam to be double the Newtonian value.

    But anyway, I think the main conclusion of this thread is that the OP needs to stop guessing based on flawed analogies and learn a bit about the specifics of GR.
     
  19. Jul 6, 2016 #18

    A.T.

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    And when you transport a gyroscope around that ring, will it's axis orientation change by a different amount, than the above angle?
     
  20. Jul 6, 2016 #19

    Jonathan Scott

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    If you go slowly enough, I think it's the same thing, although I don't have my notes to check that right now (the general formula can be found in Ciufolini & Wheeler "Gravitational and Inertia). If you go faster, that definitely affects Thomas precession and acceleration terms; the most familiar result is for orbital speed.
     
  21. Jul 6, 2016 #20

    PeterDonis

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    So we are talking an elliptical orbit, not a circular orbit, correct?
     
  22. Jul 6, 2016 #21

    Jonathan Scott

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    I was assuming approximately circular, so r is the same all the way round. For a weak field, the angle discrepancy is tiny anyway. This is the same thing as is described by the ##\gamma## parameter in the PPN approximation, giving the spatial curvature.

    Obviously the perihelion precession calculation requires an orbit which is at least slightly elliptical in order to have an effect, but that doesn't affect the general theoretical result.
     
  23. Jul 6, 2016 #22
    Say me, what is incorrect?

    We observe the angles directly - in the space!
    There no time element has any impact.
    The observation is instant: t = const, thus any angular shift must be a pure and static geometry relation, nothing more.

    Oh! I discovered an additional fantastic question:
    the time delay in GR, called sometimes as the Shapiro delay, is due to:
    1. the distance is bigger to the Sun, because the space is curved
    2. the light speed changes, and it is less than c near the Sun, thus the measured time delay?
     
  24. Jul 6, 2016 #23

    PeterDonis

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    Well, if ##r## is exactly constant, then there is zero precession and zero angle deficit. So there must be some kind of perturbation expansion about the constant ##r## orbit solution.

    You mentioned Cuifolini and Wheeler's book earlier. Is this particular angle deficit formula given there? If so, can you give a chapter/page reference?
     
  25. Jul 6, 2016 #24

    Dale

    Staff: Mentor

    This is incorrect. Often the angles are observed over time, such as when the star is "behind" the sun compared to a different time of year when the star is not behind the sun. Or precession angles before and after some period of time. Time is often an important part of the observation itself.

    Even when an observation is done at a single time, this statement is incorrect. The curvature of a ray of light requires space and time curvature even if you merely observe it at one instant.

    This is incorrect in general.

    And this is also incorrect. Even if the spacetime is static or stationary it does not imply that all curvature effects can be reduced to a purely spatial geometry.

    I remind you about the rules prohibiting personal speculation and requiring all posts to be consistent with mainstream professional science.
     
    Last edited: Jul 6, 2016
  26. Jul 6, 2016 #25

    Jonathan Scott

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    The original post was about the suggestion that the curvature of space alone produces the observed perihelion precession as an angular deficit, which I assume is understood to be clearly false.

    In the PPN formalism, the gamma term representing the curvature of space is only one part of the effect. I have been pointing out that the curvature of space (relative to a suitable coordinate system) and the effect it has on the angle of orientation (say of a protractor) around a static spatial circuit is clearly not equal to the perihelion precession. It is not entirely clear to me whether this relates to what the original post was suggesting, but other senses of angular defect such as relating circumference to radius depend on the coordinate system in a more complicated manner.

    The Ciufolini and Wheeler formula which I had in mind is equation 3.4.38 (in either form). It came up only because of the reference to a gyroscope, and is the general formula for the rate of change of orientation.
    [Edited to add accidentally omitted quote]
     
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