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felicja
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I think this should be equal to the famous precession angle, but with a negative sign:
##d\phi = 6\pi m/r##
correct?
##d\phi = 6\pi m/r##
correct?
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##.felicja said:I think this should be equal to the famous precession angle, but with a negative sign:
##d\phi = 6\pi m/r##
correct?
If you are referring to planetary orbits in the Schwarzschild vacuum, then the exact solution is in terms of the Weierstrass ##\wp## function.felicja said:...
So, the orbital precession must be equal to the angle deficit along any closed loop in the curved space, and exactly, not other.
m is not as straight forward in General Relativity as it is in Newtonian gravity.felicja said:And what is the Gaussian curvature of the space in the GR: ##K=m/r^3## ?
felicja said:And what is the Gaussian curvature of the space in the GR: ##K=m/r^3## ?
This is incorrect. GR claims that spacetime is curved in general. Furthermore, the foliation into space and time is not unique.felicja said: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
felicja said:what is the angular deficit in the GR curved space (gravitationally)?
felicja said:We observe these all effects in the space directly, not a time is relevant there!
OK, that's a third type of precession angle!A.T. said:Maybe the question is this: Would the orientation of a gyroscope precess at the same rate as it's non-circular orbit does?
Jonathan Scott said:The angle defect for a set of static rulers around the path of an orbit
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.PeterDonis said: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?
And when you transport a gyroscope around that ring, will it's axis orientation change by a different amount, than the above angle?Jonathan Scott said: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.
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.A.T. said:And when you transport a gyroscope around that ring, will it's axis orientation change by a different amount, than the above angle?
Jonathan Scott said: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
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.PeterDonis said:So we are talking an elliptical orbit, not a circular orbit, correct?
Dale said: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
Jonathan Scott said:I was assuming approximately circular, so r is the same all the way round
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.felicja said:Say me, what is incorrect?
We observe the angles directly - in the space!
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.felicja said:There no time element has any impact.
This is incorrect in general.felicja said:The observation is instant: t = const,
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.felicja said:thus any angular shift must be a pure and static geometry relation, nothing more.
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.PeterDonis said: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?
Jonathan Scott said: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.
Jonathan Scott said:The Ciufolini and Wheeler formula which I had in mind is equation 3.4.38
Angular defect, as I understand it, is a property of the spatial geometry, not of the orbit.PeterDonis said:Well, if r is exactly constant, then there is zero precession and zero angle deficit.
Light doesn't propagate instantaneouslyfelicja said: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.
It's due to space-time geometry, which also results in 1 and 2.felicja said: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?
It is difficult to compare the GR case with a classical angular defect, but I think that the model I used is the best one can do. That is, conceptually create a static frame of massless rigid rulers and protractors around the circle, go around it slowly counting up the angle and note the difference from 2 pi radians when reaching the original location (or equivalently note the distance from the original location as a fraction of the circle when a deviation of 2 pi radians is reached). Note that this result is in terms of local observations and does not depend on the choice of coordinate system, although the method of calculating it would obviously make use of a coordinate system.A.T. said:Angular defect, as I understand it, is a property of the spatial geometry, not of the orbit.
https://en.wikipedia.org/wiki/Angular_defect
An angle deficit in the curved space in General Relativity (GR) refers to the phenomenon where the sum of angles of a triangle in a curved space is less than 180 degrees. This is caused by the curvature of space, which is a fundamental concept in GR. In other words, an angle deficit is a measure of the deviation from Euclidean geometry in a curved space.
Yes, an angle deficit can be observed in real life. One example is the observation of light bending around massive objects, such as stars, due to the curvature of space. This was first predicted and observed by Einstein's theory of General Relativity.
An angle deficit is a consequence of the theory of relativity, specifically General Relativity. This theory describes how gravity is not a force between masses, but rather a result of the curvature of space and time caused by massive objects. The presence of mass causes a distortion in the geometry of space, leading to an angle deficit in curved space.
No, an angle deficit and gravitational lensing are not the same. While both are consequences of the curvature of space, gravitational lensing refers to the bending of light around massive objects, while an angle deficit is a measure of the deviation from Euclidean geometry in a curved space.
Yes, an angle deficit can be calculated using mathematical formulas derived from the theory of General Relativity. These formulas take into account the mass and distance of the object causing the curvature of space. In some cases, the angle deficit can also be measured directly through observations of the bending of light or other phenomena affected by the curvature of space.