Thank @KyleKanos for a suggestion, I am rephrasing this question to hopefully make it more clear.

According to this publication in Physics Letters, 2000: Aberration and the Speed of Gravity:

It is well known that if a charged source moves at a constant velocity, the electric field experienced by a test particle points toward the source’s “instantaneous” position rather than its retarded position.

The paper then extends this result to gravity:

aberration in general relativity is almost exactly canceled by velocity-dependent interactions

More specifically:

the gravitational acceleration is directed toward the retarded position of the source quadratically extrapolated toward its “instantaneous” position, up to small nonlinear terms and corrections of higher order in velocities.

The author immediately clarifies:

Does [this] imply that gravity propagates instantaneously? As in the case of electromagnetism, it clearly does not.

Following the logic in the paper, the Earth is attracted to the instantaneous position of the Sun 8 minutes or 4 Sun's diameters ahead of its observed retarded position in the sky.

The paper also explains that this conclusion is precise only to the radiative term. In case of the Earth rotating around the Sun, this correction is very small, because the gravitational radiation emitted by the Earth does not significantly change its orbit. (The situation, of course, would be very different in a strong gravity of a neutron star or black hole.)

My question is on the meaning of "toward the instantaneous position" in differential geometry. For example, in the Schwarzschild spacetime, does the gravitational acceleration of the Earth point in the direction of the spacelike geodesic between the Earth and the Sun? If not, then what is the mathematical definition of the direction "toward the instantaneous position" described in the paper?

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    $\begingroup$ FWIW, the current version seems to be a perfectly good question. I don't know whether it's similar enough to the original to be an edit, or whether it should be posted as a new question. $\endgroup$ – Javier Jul 23 '19 at 18:48
  • $\begingroup$ @Javier It is exactly the same question. It doesn't ask any more or any less than before, just phrased in such a way that my detailed explanation of what I am asking cannot be misinterpreted by some as "a personal theory". If you believe my edit is helpful and the question is perfectly good, please upvote to cancel the downvotes that are no longer relevant. Thanks! $\endgroup$ – safesphere Jul 23 '19 at 19:16
  • $\begingroup$ A conversation which was largely about a previous version of this question has been moved to chat. $\endgroup$ – rob Jul 26 '19 at 23:29

You ask "Does such a 'path of gravity' make sense?"

The answer is no, for a couple of reasons.

Here is how I interpret your "path of gravity." Construct a curve such that the tangent vector at each point is in the direction of the force of gravity at that point. In Newtonian gravity with the Sun at rest this path would simply follow radial lines inward to the origin.

In general relativity the construction does not work because the force of gravity depends on the speed of the object under consideration. Explicitly, the geodesic equation is $$ m \frac{d^2x^\mu}{d\tau^2} = -m \Gamma^\mu_{\rho\sigma} \frac{dx^\rho}{d\tau}\frac{dx^\sigma}{d\tau} $$ and in situations with non-trivial gravity $\Gamma^\mu_{ij}$ and $\Gamma^\mu_{it}$ may be non-zero and so there are factors of $dx^i/d\tau \propto v^i$ in the force. Hence, there is no unique gravitational force vector at each point and so we cannot define a unique path of gravity in this way.

Another way of seeing that this construction doesn't tell us anything very useful is to consider freely-falling reference frames. One of the fundamental ideas of general relativity is that the laws of physics must be in terms of objects that exist in all coordinate systems. It is also a fundamental principle of general relativity that in freely-falling reference frames there is no local gravitational force! In these coordinate systems we therefore cannot construct a path of gravity. And hence, even if we could construct the path of gravity in some coordinate system, it cannot be a result of the fundamental laws of physics. At best it could be a neat trick of a particular coordinate system in a particular situation, but nothing deeper.

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  • $\begingroup$ Thank you for the answer. You present a general case, including "non-trivial gravity", and make a conclusion of no general result, "even if we could construct the path of gravity in some coordinate system". However, the question is specifically about an established system without "non-trivial gravity" and without aberrations. The question pretty much is about the Earth rotating around the Sun forever with everything else (including the mass of the Earth) ignored. Can you please add the specific case of the Schwarzschild spacetime to your answer? Is the path in question a radial geodesic? $\endgroup$ – safesphere Jul 22 '19 at 16:24
  • $\begingroup$ The whole point of the answer is that the path does not exist the way you define it, even in the Schwarzschild metric in the standard coordinates. It does not exist because there is no unique gravitational potential because there is no unique gravitational force at each point in space. So how are we supposed to trace out your path? $\endgroup$ – Luke Pritchett Jul 22 '19 at 16:26
  • $\begingroup$ It is often the case that specific helpful mental pictures or concepts fail to generalize from Newtonian physics to modern physics. That seems to be the case here. All one can do is shrug and try and find a new way of understanding what's going on. Best of luck. $\endgroup$ – Luke Pritchett Jul 22 '19 at 16:36
  • $\begingroup$ Not Newtonian since worldlines there are very different. I was thinking of the Schwarzschild geometry instead.I'll let you know after some time. Thanks again! $\endgroup$ – safesphere Jul 22 '19 at 17:09
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    $\begingroup$ That paper says that if you have a moving gravitational source then (in a particular change of coordinates) the acceleration of a test particle points towards the "current" position of the source, not the retarded position. That "current" position is not technically the current position, but actually the retarded position extrapolated forward in time by a constant-acceleration approximation. This is according to GR. The larger point of the paper seems to the be that you can't describe actual gravity by using Newtonian physics with a speed-of-light delay added in by hand. $\endgroup$ – Luke Pritchett Jul 23 '19 at 16:08

Although the sun is moving at high speed in its journey around the galaxy, from our point of view it is at rest and stays in the same place. As Einstein said, all motion is relative. Gravity emanates from it at the same speed as light, and travels the same infinite distance. Therefore there is no mismatch between where we see the sun and where its gravity comes from. If by some miraculous means the gravitational field emanating from the sun vanished at this moment, we would know nothing about it for over 8 minutes.

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  • $\begingroup$ You are mixing the concepts of "field" and "wave". See for example this Q&A: "Once you factor in everything you need to in order to model a real system behaving in a realistic manner, you find that all the aberrations you might expect because of a finite speed of light end up canceling out, so gravity acts like it's instantaneous, even though the underlying phenomenon is most definitely not." - physics.stackexchange.com/questions/5456/… $\endgroup$ – safesphere Jul 21 '19 at 19:05
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    $\begingroup$ From my point of view, the Sun moves around a lot. $\endgroup$ – G. Smith Jul 21 '19 at 19:09
  • $\begingroup$ @G. Smith : I can assure you, the Earth goes round the Sun. $\endgroup$ – John Duffield Jul 29 '19 at 19:36

You are asking if the gradient of the potential follows a spacelike geodesic.

I just answered a question:

Let's say that the Sun moves away from us, this should cause a change in gravitational force, when will this change be noticed by us?

Now it is very important to clarify that you are not asking about GWs. You are asking about the static gravitational field of the Sun.

And you are asking if the Sun as the center of the Solar system (center of gravity), has a gravitational potential, where the gradient pointing towards the Sun will (from Earth in your case) will be a spacelike geodesic.

I believe you are asking if it is a spacelike worldline.

space-like curves falling outside the light cone. Such curves may describe, for example, the length of a physical object. The circumference of a cylinder and the length of a rod are space-like curves.

Now the Solar system, if we look at it as a reference frame, is moving through space.

You are basically asking, if a change in the static gravitational field, would be felt in the frame of the Solar system, at a certain speed.

Now in reality we do not have a accepted theory of quantum qravity. We do not really know how the static gravitational field works, we have theories, that describe the data from experiments.

Now you are basically asking if the Sun's or the Earth's gravitational field (which both effect each other) changed, how fast would that be felt by the other object in the frame of the Solar system.

This is where the confusion is.

In classical theories of gravitation, the changes in a gravitational field propagate. A change in the distribution of energy and momentum of matter results in subsequent alteration, at a distance, of the gravitational field which it produces. In the relativistic sense, the "speed of gravity" refers to the speed of a gravitational wave, which, as predicted by general relativity and confirmed by observation of the GW170817 neutron star merger, is the same speed[1] as the speed of light (c).

Now this is talking about any change traveling at the speed of light, that is, that would be a gradient, where the worldline would be spacelike.

The consequence of this is that static fields (either electric or gravitational) always point directly to the actual position of the bodies that they are connected to, without any delay that is due to any "signal" traveling (or propagating) from the charge, over a distance to an observer. This remains true if the charged bodies and their observers are made to "move" (or not), by simply changing reference frames. This fact sometimes causes confusion about the "speed" of such static fields, which sometimes appear to change infinitely quickly when the changes in the field are mere artifacts of the motion of the observer, or of observation.


Now this one says, that the gradient must point towards the center of gravity always. Changes (position) in the gravitational field of the Solar system (as a whole reference frame, including the Sun and the Earth), would be felt inside the Solar system instantly. This would require the gradient to be on a lightlike worldline. The Sun and the Earth are traveling through space inside the gravitational field of the Solar system as a common system.

Now you are basically asking, if the trajectory of the Sun (or Earth) would change because of an external influence, how fast would this change in the gravitational field be felt by the other object inside the reference frame (inside the Solar system).

Basically, if the Sun's trajectory would be altered (because another object would hit the Sun), then the Sun's gravitational field (without the Earth's) would change from its normal trajectory, and the question is, would these changes on Earth be felt instantly or just after 8 minutes.

In reality we use virtual particles when we describe the gravitational field. Why? Because in reality we do not know how it works on the quantum level. These virtual gravitons are a mathematical model. These particles do not obey SR, and do not obey the speed of light.

You are asking whether the gradient of the Sun's static gravitational field would instantly change to the Sun as the Sun would alter its trajectory.

This would mean, that though the Sun is altering its original trajectory, the distance between the Sun and Earth (the Earth orbit around the Sun) would not change.

So basically the whole Solar system as a common reference frame would alter its trajectory together (though only the Sun was hit by another object).

The static field is described by virtual particles, because they do not obey the speed of light. As the Sun would be altering its trajectory, the gradient would too, moving together with the Sun. The change in the direction of the gradient is what would change instantly, but actually nothing in this case is traveling faster then c. No information is traveling faster then c. The static field is already there, inside it the Earth, and so from the Earth the gradient would keep pointing towards the Sun.

You are asking about towards instantaneous position. The gradient is pointing towards the Sun, and towards its instantaneous position.

If you change the trajectory of the Sun, nothing is moving faster then c. The static field is described in math with virtual particles. The gradient point towards the instantaneous position of the Sun.

There is no contradiction with SR. From Earth, the Sun's altering its original trajectory would be seen as a change in the gradient's direction. It would always point towards the Sun. The gradient's direction's change would not be delayed with 8 minutes.

We do not have a accepted quantum theory of gravity, thus, if you are asking how it is possible that the gravitons' path (gradient) would change, the answer is, those are virtual gravitons. The path of virtual gravitons is basically what you are asking. Virtual particles follow faster then light geodesics (worldlines).

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