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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?
     
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